MATHS 5 Marks Questions

If the the expansion of $(\text{x}+\text{a})^{\text{n}},$ the 2^nd, 3^rd and 4^th are ${^\text{n}}\text{C}_{\text{1}}\text{x}^{\text{n}-1}\text{a}^{1}, \ {^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}$ and,

According to the quation, 

 ${^\text{n}}\text{C}_{\text{1}}\text{x}^{\text{n}-1}\text{a}^{1}=240$

${^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}=720$

${^\text{n}}\text{C}_{\text{3}}\text{x}^{\text{n}-3}\text{a}^{3}=1080$

Now,

$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}}{{^\text{n}}\text{C}_{\text{3}}\text{x}^{\text{n}-3}\text{a}^{3}}=\frac{720}{240}$

$\Rightarrow \frac{\text{n}-1}{2\text{x}}\text{a}=3$

$\Rightarrow \frac{\text{a}}{\text{x}}=\frac{6}{\text{n}...1}\ ...(\text{i})$

Also,

$\frac{{^\text{n}}\text{C}_{\text{3}}\text{x}^{\text{n}-3}\text{a}^{3}}{{^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}}=\frac{1080}{720}$

$\Rightarrow \frac{\text{n}-2}{3\text{x}}\text{a}=\frac{3}{2}$

$\Rightarrow \frac{\text{a}}{\text{x}}=\frac{9}{2\text{n}...4}\ ...(\text{ii})$

From (i) and (ii), we get

$\frac{6}{\text{n}-1}=\frac{9}{2\text{n}-4}$

$\Rightarrow \text{n}=5$

Putting in equation (i), we get

$\Rightarrow 2\text{a}=3\text{x}$

Now,

${^\text{5}}\text{C}_{\text{1}}\text{x}^{\text{5}-1}\big(\frac{\text{3}}{2}\text{x}\big)=240$

$\Rightarrow 15\text{x}^{5}=480$

$\Rightarrow \text{x}^{5}=32$

$\Rightarrow \text{x}=2$

By putting the value of x and n in (i). we get

$\text{a}=3$

$\Big\{\text{a}^2+\sqrt{\text{a}^2-1}\Big\}^4+\Big\{\text{a}^2-\sqrt{\text{a}^2-1}\Big\}^4$

$\text{Let}\ \text{a}^2=\text{A},\ \sqrt{\text{a}^2-1}=\text{B}$

$(\text{A}+\text{B})^4+(\text{A}-\text{B})^4$

$=\text{B}^4+{^4\text{C}}_1\text{AB}^3+{^4\text{C}}_2\text{A}^2\text{B}^2+{^4\text{C}}_3\text{A}^3\text{B}+\text{A}^2+\text{B}^4\\-{^4\text{C}}_1\text{AB}^3+{^4\text{C}}_2\text{A}^2\text{B}^2-{^4\text{C}}_3\text{A}^3\text{B}+\text{A}^4$

$=2\big(\text{A}^4+{^4\text{C}}_2\text{A}^2\text{B}^2+\text{B}^4\big)$

$=2\big(\text{A}^2+6\text{A}^2\text{B}^2+\text{B}^4\big)$

$=2\Big(\text{a}^8+6\text{a}^4(\text{a}^2-1)+(\text{a}^2-1)^2\Big)$

$=2\Big[\text{a}^8+6\text{a}^6-6\text{a}^4+\text{a}^4+1-2\text{a}^2\Big]$

$\Big\{\text{a}^2+\sqrt{\text{a}^2-1}\Big\}+\Big\{\text{a}^2-\sqrt{\text{a}^2-1}\Big\}$

$=2\text{a}^8+12\text{a}^6-10\text{a}^4-4\text{a}^4+2$

$3^{3\text{n}}-26\text{n}-1$

$=(3^3)^\text{n}-26\text{n}-1$

$=27^\text{n}-26\text{n}-1$

$=(1+26)^\text{n}-26\text{n}-1$

$\Big({^\text{n}\text{C}}_0+{^\text{n}\text{C}}_1(26)^1+{^\text{n}\text{C}}_2(26)^2+.....+676(26)^{\text{n}-2}\Big)-26\text{n}-1$

$=676\Big({^\text{n}\text{C}}_2+......+(26)^{\text{n}-2}\Big)$

$\therefore3^{3\text{n}}-26\text{n}-1$ is divisible for $\text{n}\in\text{N}.$

Hence, proved

$(1+\text{x}+2\text{x}^{3})\Big(\frac{3}{2}\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{9}$

$(1+\text{x}+2\text{x}^{3})\bigg[\Big(\frac{3}{2}\text{x}^{2}\Big)^{9}-{^\text{9}}\text{C}_{\text{1}}\Big(\frac{3}{2}\text{x}^{2}\Big)^{8}\ \frac{1}{3\text{x}}.....\\ +{^\text{9}}\text{C}_{\text{6}}\Big(\frac{3}{2}\text{x}^{2}\Big)^{3}\Big(\frac{1}{3\text{x}}\Big)^{6}-{^\text{9}}\text{C}_{\text{7}}\Big(\frac{3}{2}\text{x}^{2}\Big)^{2}\Big(\frac{1}{3\text{x}}\Big)^{7}\bigg]$

In the second bracket, we have to search the term so x⁰ and $\frac{1}{\text{x}^{3}}$ which when multiplying by 1 and 2x³ is first bracket will give the term independent of x. The term containing $\frac{1}{\text{x}}$ will not occur is second bracket.

The teram independent of x,

$=1\Big[{^\text{9}}\text{C}_{\text{6}}\frac{3^{3}}{2^{2}}\times\frac{1}{3^{6}}\Big]-2\text{x}^{3}\Big[{^\text{9}}\text{C}_{\text{7}}\frac{3^{3}}{2^{2}}\times\frac{1}{3^{7}}\times\frac{1}{\text{x}^{3}}\Big]$ 

$=\Big[\frac{9\times8\times7}{1\times2\times3}\times\frac{1}{8\times27}\Big]-2\Big[\frac{9\times8}{1\times2}-\frac{1}{4\times243}\Big]$

$=\frac{7}{18}-\frac{2}{27}$

$=\frac{17}{54}$

In the binomial expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ there are (n + 1) terms.

The third term from the end in the expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$  is the third term from the beginning in the expansion of  $\Big(\text{x}^{\frac{1}{3}}+\text{y}^{\frac{1}{2}}\Big)^{\text{n}}.$

The binomial coefficient of the third term from the end is 45.

$\therefore {^\text{n}}\text{C}_{\text{2}}=45$

$\Rightarrow \frac{\text{n}(\text{n}-1)}{2}=45$

$\Rightarrow \text{n}^{2}-\text{n}-90=0$

$\Rightarrow (\text{n}-10)(\text{n}+9)=0$

$\Rightarrow \text{n}=10$

Let T₆ be the six^th term in the binomoal expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}}.$ 

Then,

$\text{T}_{6}={^\text{n}}\text{C}_{\text{5}}\Big(\text{y}^{\frac{1}{2}}\Big)^{\text{n}-5}\Big(\text{x}^{\frac{1}{3}}\Big)^{5}$

$={^\text{10}}\text{C}_{\text{5}}\text{y}^{\frac{5}{2}}\text{x}^{\frac{5}{3}}$

$=252\text{y}^{\frac{5}{2}}\text{x}^{\frac{5}{3}}$

Hence, the six^th term in the expansion of $\Big(\text{y}^{\frac{1}{2}}+\text{x}^{\frac{1}{3}}\Big)^{\text{n}},$ is $252\text{y}^{\frac{5}{2}}\text{x}^{\frac{5}{3}}.$

We have,

$(98)^5=(100-2)^5$

$={^5\text{C}}_0\times100^5+{^5\text{C}}_1\times100^4\times(-2)+{^5\text{C}}_2\times100^3\times(-2)^2\\+{^5\text{C}}_3\times100^2\times(-2)^3+{^5\text{C}}_4\times100\times(-2)^4+{^5\text{C}}_5\times(-2)^5$

$={^5\text{C}}_0\times100^5-{^5\text{C}}_1\times100^4\times2+{^5\text{C}}_2\times100^3\times4\\-{^5\text{C}}_3\times100^2\times8+{^5\text{C}}_4\times100\times16-{^5\text{C}}_5\times32$

$=100^5-10\times100^4+40\times100^3-80\times100^2+80\times100-32$

$=10000000000-1000000000+40000000-800000+8000-32$

$=10040008000-1000800032$

$=9039207968$

$\therefore(98)^5=9039207968$

$\Big(2\text{x}-\frac{\text{x}^{2}}{4}\Big)^{9}$

Here, n is an odd number.

Therefore, the middle terms are $\Big(\frac{\text{n+1}}{2}\Big)$ and $\Big(\frac{\text{n+1}}{2}+1\Big)$ 5th and 6th terms.

Now, we have,

$\text{T}_{5}=\text{T}_{4+1}$

$={^\text{9}}\text{C}_{\text{4}}(2\text{x})^{9-4}\big(\frac{-\text{x}^{2}}{4}\big)^{4}$

$=\frac{9\times8\times7\times6}{4\times3\times2}\times2^{5}\frac{1}{4^{4}}\text{x}^{5+8}$

$=\frac{63}{4}\text{x}^{13}$

T₆ = a

T₇ = b

T₈ = c

T₉ = d

We have, (x + 1)ⁿ

Then,

$\frac{\text{T}_{\text{r}+1}}{\text{T}_{\text{r}}}=\frac{\text{n-r}+1}{\text{r}}.\frac{1}{\text{x}}$

Put r = 6,

$\frac{\text{T}_7}{\text{T}_6}=\frac{\text{n}-6+1}{6}.\frac{1}{\text{x}}=\frac{\text{n}-5}{6}.\frac{1}{\text{x}}=\frac{\text{b}}{\text{a}}\ \dots(\text{a})$

Put r = 7

$\frac{\text{T}_8}{\text{T}_7}=\frac{\text{n}-7+1}{7}.\frac{1}{\text{x}}=\frac{\text{n}-6}{7}.\frac{1}{\text{x}}=\frac{\text{c}}{\text{b}}\ \dots(\text{b})$

Put r = 8

$\frac{\text{T}_9}{\text{T}_8}=\frac{\text{n}-8+1}{8}.\frac{1}{\text{x}}=\frac{\text{n}-7}{7}.\frac{1}{\text{x}}=\frac{\text{d}}{\text{c}}\ \dots(\text{c})$

Now,

$\frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}=\frac{4\text{a}}{3\text{c}}$

Divide by bc in L.H.S, we get

$\Rightarrow \frac{\frac{\text{b}^{2}}{\text{bc}}-\frac{\text{ac}}{\text{bc}}}{\frac{\text{c}^{2}}{\text{bc}}-\frac{\text{bd}}{\text{bc}}}=\frac{4\text{a}}{3\text{c}}$

$\Rightarrow \frac{\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{b}}}{\frac{\text{c}}{\text{b}}-\frac{\text{d}}{\text{c}}}=\frac{4\text{a}}{3\text{c}}$

Take L.H.S

By Equation (a), (b), (c)

$\Rightarrow\ \frac{\frac{7\text{x}}{\text{n}-6}-\frac{6\text{x}}{\text{n}-5}}{\frac{\text{n}-6}{7\text{x}}-\frac{\text{n}-7}{8\text{x}}}$

$\Rightarrow\ \frac{\text{x}\Big[\frac{7}{\text{n}-6}-\frac{6}{\text{n}-5}\Big]}{\frac{1}{\text{x}}\Big[\frac{\text{n}-6}{7}-\frac{\text{n}-7}{8}\Big]}$

$\Rightarrow\ \text{x}^2\Bigg[\frac{\text{n}-35-6\text{n}+36}{\big(\text{n}-6\big)\big(\text{n}-5\big)}\Bigg]\times\frac{56}{8\text{n}-48-7\text{n}+49}$

On solving this, we get

$\frac{56\text{x}^2}{\big(\text{n}-6\big)\big(\text{n}-5\big)}\ \dots(\text{X})$

On Taking R.H.S

$\Rightarrow\ \frac{4\text{a}}{3\text{c}}=\frac{4}{3}\Big(\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{c}}\Big)$

On putting the values from equation (a), (b), (c)

$\Rightarrow\ \frac{4}{3}\Big[\frac{6\text{x}}{\text{n}-5}\times\frac{7\text{x}}{\text{n}-6}\Big]$

$\frac{56\text{x}^2}{\big(\text{n}-6\big)\big(\text{n}-5\big)}\ \dots(\text{Y})$

By equation X and Y, it is proved that

L.H.S = R.H.S

We have,

$(1+2\text{a})^{4}(2-\text{a})^{5}$

$=\Big[{^\text{4}}\text{C}_{\text{0}}\big(2\text{a}\big)^{0}+{^\text{4}}\text{C}_{\text{1}}\big(2\text{a}\big)^1+{^\text{4}}\text{C}_{\text{2}}\big(2\text{a}\big)^{2}+{^\text{4}}\text{C}_{\text{3}}\big(2\text{a}\big)^{3}+{^\text{4}}\text{C}_{\text{4}}\big(2\text{a}\big)^{4}\Big]\\ \times\Big[{^\text{5}}\text{C}_{\text{0}}(2)^{5}(-\text{a})^{0}+{^\text{5}}\text{C}_{\text{1}}(2)^{4}(-\text{a})^{1}+{^\text{5}}\text{C}_{\text{2}}(2)^{3}(-\text{a})^{2}+{^\text{5}}\text{C}_{\text{2}}(2)^{2}(-\text{a})^{3}+{^\text{5}}\text{C}_{\text{1}}(2)^{1}(-\text{a})^{5}$

$=\Big[1+8\text{a}+24\text{a}^{2}+32\text{a}^{3}+16\text{a}^{4}\Big]\times\Big[32-80\text{a}+80\text{a}^{2}-40\text{a}^{3}+10\text{a}^{4}-\text{a}^{5}\Big]$

Coefficient of $\text{a}^{4}=10-320+1920-2560+512=-438$

We have,

$(102)^5=(100+2)^5$

$={^5\text{C}}_0\times100^5+{^5\text{C}}_1\times100^4\times2+{^5\text{C}}_2\times100^3\times2^2\\+{^5\text{C}}_3\times100^2\times2^3+{^5\text{C}}_4\times100\times2^4+{^5\text{C}}_5\times2^5$

$=100^5+5\times100^4\times2+10\times100^3\times2^2\\+10\times100^2\times2^3+5\times100\times2^4+2^5$

$=10000000000+1000000000+40000000+800000+8000+32$

$=11040808032$

$\therefore(102)^5=11040808032$

We have,

$(1+\text{x})^{\text{n}}$

Suppose r, (r + 1) and (r + 2) are three consecutive terms in the given expansion.

The coefficirnts of these terms are ${^\text{n}}\text{C}_{\text{r}-1},{^\text{n}}\text{C}_{\text{r}}, {^\text{n}}\text{C}_{\text{r}+1}.$

Coefficients of rth terms $={^\text{n}}\text{C}_{\text{r}-1}=76$

Coefficients of (r + 1) terms $={^\text{n}}\text{C}_{\text{r}}=95$

Coefficients of (r + 2) terms $={^\text{n}}\text{C}_{\text{r}-1}=76$

$\Rightarrow {^\text{n}}\text{C}_{\text{r}-1}={^\text{n}}\text{C}_{\text{r}+1}$

$\Rightarrow \text{r}-1+\text{r}+1=\text{n}$

$\Rightarrow \text{r}=\frac{\text{n}}{2}$

Now,

$\therefore \frac{{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{96}{76}$

$\Rightarrow \frac{\text{n}-\text{r}+1}{\text{r}}=\frac{95}{76}$

$\Rightarrow \frac{\frac{\text{n}}{2}+1}{\frac{\text{n}}{2}}=\frac{95}{76}$

$\Rightarrow 38\text{n}+76=\frac{95\text{n}}{2}$

$\Rightarrow \frac{19\text{n}}{2}=76$

$\Rightarrow \text{n}=8$

We have,

$(1+\text{x})^{\text{n}}$

Coefficient of pth term $={^\text{n}}\text{C}_{\text{p}-1}$

Coefficient of qth term $={^\text{n}}\text{C}_{\text{q}-1}$

It is given that, these coefficients are equal.

${^\text{n}}\text{C}_{\text{p}-1}={^\text{n}}\text{C}_{\text{q}-1}$

$\Rightarrow \text{p}-1=\text{q}-1$ or $\text{p}-1+\text{q}-1=\text{n}$

$\Rightarrow \text{p}-\text{q}=0$ or $\text{p}+\text{q}=\text{n}+2$

$\therefore\ \text{p}+\text{q}=\text{n}+2$

Hence proved.

In the binomial expansion of $\Big(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}},\big[(\text{n}+1)-7+1\big]^\text{th}\text{i.e.,}\ (\text{n}-5)^{\text{th}}$ term from the beginning is the 7^th term from the end.

Now,

$\text{T}_{7}={^\text{n}}\text{C}_{\text{6}}\Big(\sqrt[3]{2}\Big)^{\text{n}-6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{6}={^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}$

And,

$\text{T}_{\text{n}-5}={^\text{n}}\text{C}_{\text{n}-6}\Big(\sqrt[3]{2}\Big)^{6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}-6}={^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}$

It is given that,

$\text{T}_{7}=\text{T}_{\text{n}-5}$

$\Rightarrow {^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}$

$\Rightarrow {^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}$

$\Rightarrow \frac{2^{\frac{\text{n}}{3}-2}}{2^{2}}=\frac{3^{2}}{3^{\frac{\text{n}}{3}-2}}$

$\Rightarrow (6)^{\frac{\text{n}}{3}-2}=6^{2}$

$\Rightarrow \frac{\text{n}}{3}-2=2$

$\Rightarrow \text{n}=12$

Hence, the value of n is 12.

We have,

$\Big(\frac{3}{2}\text{x}^{2}-\frac{1}{3\text{x}}\Big)^{6}$

In expansion,

$\text{T}_{\text{r+1}}= {^\text{6}}\text{C}_{\text{r}}\Big(\frac{3\text{x}^{2}}{2}\Big)^{6-\text{r}}\Big(-\frac{1}{3\text{x}}\Big)^{\text{r}}$

$= {^\text{6}}\text{C}_{\text{r}}\Big(\frac{3}{2}\Big)^{6-\text{r}}(\text{x}^{12-3\text{r}})\Big(-\frac{1}{3}\Big)^{\text{r}}$

Let T_r+1 be independent of x,

$12-3\text{r}=0$

$\text{r}=4$

Required term

$\text{T}_{\text{r+1}}= \text{T}_{\text{4+1}}=\text{T}_{\text{5}}={^\text{6}}\text{C}_{\text{r}}\Big(\frac{3}{2}\Big)^{6-\text{4}}\Big(\frac{-1}{3}\Big)^{4}\text{x}^{12-3(4)}$

$=15\Big(\frac{9}{4}\Big)\Big(\frac{1}{81}\Big)\text{x}^{0}$

$=\frac{5}{12}$

We have,

$\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}+\text{x}^{-\frac{1}{5}}\Big)^{8}$

Let (r + 1)^th term be independent of x.

$\text{T}_{\text{r}+1}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-\text{r}}\Big(\text{x}^{\frac{-1}{5}}\Big)^{8}$

 $={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\Big(\text{x}^\frac{1}{3}\Big)^{8-\text{r}}\times\Bigg(\frac{1}{\text{x}^{\frac{1}{5}}}\Bigg)^{\text{r}}$

$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x}\big)^\frac{8-\text{r}}{3}\times\Bigg(\frac{1}{\text{x}^{\frac{1}{5}}}\Bigg)$

$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x})^{\frac{8-\text{r}}{3}-\frac{\text{r}}{5}}$

$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x})^{\frac{40-5\text{r}-3\text{r}}{15}}$

$={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\Big)^{8-\text{r}}\times\big(\text{x})^{\frac{40-8\text{r}}{15}}$

If it is independent of x, We must have

$\Rightarrow \frac{40-8\text{r}}{15}=0$

$\Rightarrow 8\text{r}=40$

$\Rightarrow \text{r}=5$

The term independet of x,

Now,

$\text{T}_{6}={^\text{8}}\text{C}_{\text{r}}\Big(\frac{1}{2}\text{x}^{\frac{1}{3}}\Big)^{8-5}\Big(\text{x}^{\frac{-1}{5}}\Big)^{5}$

$=56\times\Big(\frac{1}{2}\Big)^{3}$

$=56\times\frac{1}{8}$

$=7$

Suppose the three consective terms are T_r-1, T_r and T_r+1.

Coefficients of these terms are $​{^\text{n}}\text{C}_{\text{r}-2}, ​{^\text{n}}\text{C}_{\text{r}-1}$ and $​{^\text{n}}\text{C}_{\text{r}},$ respectively.

These corfficients are equal to 220, 495 and 792.

$\therefore \frac{​{^\text{n}}\text{C}_{\text{r}-2}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{220}{495}$

$\Rightarrow \frac{\text{r}-1}{\text{n}-\text{r}+2}=\frac{4}{9}$

$\Rightarrow 4\text{n}+17=13\text{r}\ ...(\text{i})$

Also,

$\therefore \frac{​{^\text{n}}\text{C}_{\text{r}}}{{^\text{n}}\text{C}_{\text{r}-1}}=\frac{792}{495}$

$\Rightarrow \frac{\text{n}-\text{r}-1}{\text{r}}=\frac{8}{5}$

$\Rightarrow 5\text{n}-5\text{r}+5=8\text{r}$

$\Rightarrow 5\text{n}+5=13\text{r}$

$\Rightarrow 5\text{n}+5=4\text{n}+17$

$\Rightarrow \text{n}=12$

$(\sqrt{2}+1)^6+(\sqrt{2}-1)^6$

$={^6\text{C}}_0(\sqrt2)^6+{^6\text{C}}_1(\sqrt2)^5+{^6\text{C}}_2(\sqrt2)^4+{^6\text{C}}_3(\sqrt2)^3\\{^6\text{C}}_4(\sqrt2)^2+{^6\text{C}}_5(\sqrt2)+{^6\text{C}}_6+{^6\text{C}}_0(\sqrt2)^6-\\{^6\text{C}}_1(\sqrt2)^5+{^6\text{C}}_2(\sqrt2)^4-{^6\text{C}}_3(\sqrt2)^3+{^6\text{C}}_3(\sqrt2)^3-{^6\text{C}}_4(\sqrt2)^2-{^6\text{C}}_6(\sqrt2)^0$

$=2\big[2^3+15\times2^2+15\times2+1\big]$

$=2\big[8+60+30+1\big]=2(99)=198$

$(1.01)^{10}+(1-0.01)^{10}=(1-0.01)^{10}+(1-0.01)^{10}$

$=\Big({^{10}\text{C}}_1+{^{10}\text{C}}_2\frac{1}{10^2}+{^{10}\text{C}}_3\frac{1}{10^3}...+{^{10}\text{C}}_{10}\frac{1}{10^{10}}\Big)\\+\Big({^{10}\text{C}}_1-{^{10}\text{C}}_2\frac{1}{10^2}+{^{10}\text{C}}_3\frac{1}{10^3}-{^{10}\text{C}}_4\frac{1}{10^4}+....\Big)$

$=2\Big({^{10}\text{C}}_1-{^{10}\text{C}}_3\frac{1}{10^3}+{^{10}\text{C}}_5\frac{1}{10^5}+{^{10}\text{C}}_7\frac{1}{10^7}+{^{10}\text{C}}_9\frac{1}{10^9}\Big)$

$=2\Big(10+\frac{10!}{3!7!}\frac{1}{1000}+\frac{10!}{5!5!}\frac{1}{(10)^5}+\frac{10!}{7!3!}\times\frac{1}{10^7}+\frac{10!}{9!1!}\frac{1}{10^9}\Big)$

$=2\Big(10+\frac{9\times8}{3\times2\times1000}+\frac{9\times8\times7\times6}{5\times4\times3\times2\times10^5}+\frac{9\times8}{3\times2\times10^7}+\frac{1}{10^8}\Big)$

$=2.00900042$

We have,

$(1+\text{x})^{\text{n}}$

Now,

Coefficient of 5th term $={^\text{n}}\text{C}_{\text{5}-1}={^\text{n}}\text{C}_{\text{4}}$

Coefficient of 5th term $={^\text{n}}\text{C}_{\text{6}-1}={^\text{n}}\text{C}_{\text{5}}$

Coefficient of 5th term $={^\text{n}}\text{C}_{\text{7}-1}={^\text{n}}\text{C}_{\text{6}}$

It is given that these coefficients are in A.P.

$\therefore\ 2\ {^\text{n}}\text{C}_{\text{5}}={^\text{n}}\text{C}_{\text{4}}+{^\text{n}}\text{C}_{\text{6}}$

$\Rightarrow 2\Big[\frac{\text{n}!}{(\text{n}-5)!5!}\Big]=\frac{\text{n}!}{(\text{n}-4)!4!}+\frac{\text{n}!}{(\text{n}-6)!6!}$

$\Rightarrow\ \frac{2}{(\text{n}-5)!5!}=\frac{1}{(\text{n}-4)!4!}+\frac{1}{(\text{n}-6)!6!}$

$\Rightarrow \frac{2}{(\text{n}-5)(\text{n}-6)!5\times4!}=\frac{1}{(\text{4})(\text{n}-5)(\text{n}-6)!4!}+\frac{1}{(\text{n}-6)!6\times5\times4!}$

$\Rightarrow \frac{2}{(\text{n}-5)\times5}=\frac{1}{(\text{n}-4)(\text{n}-5)}+\frac{1}{6\times5}$

$\Rightarrow \frac{2}{5(\text{x}-5)}-\frac{1}{30}=\frac{1}{(\text{n}-4)(\text{n}-5)}$

$\Rightarrow \frac{12-(\text{n}-5)}{30(\text{n}-5)}=\frac{1}{(\text{n}-4)(\text{n}-5)}$

$\Rightarrow \frac{17-\text{n}}{30}=\frac{1}{(\text{n}-4)}$

$\Rightarrow 17\text{n}-68-\text{n}^{2}+4\text{n}=30$

$\Rightarrow 21\text{n}-68-\text{m}^{2}-30=0$

$\Rightarrow 21\text{n}-\text{n}^{2}-98=0$

$\Rightarrow \text{n}^{2}-21\text{n}+98=0$

$\Rightarrow \text{n}^{2}-7\text{n}-14\text{n}+98=0$

$\Rightarrow \text{n}(\text{n}-7)-17(\text{n}-7)=0$

$\Rightarrow (\text{n}-7)(\text{n}-14)=0$

$\Rightarrow \text{n}=7, \text{n}=14$

We have,

$\Big(\sqrt[3]{\text{x}}+\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{18}, \text{x}>0$

Let (r + 1)^th term be independent of x.

$\text{T}_{\text{r}+1}={^\text{18}}\text{C}_{\text{r}}\big(\sqrt[3]{\text{x}}\big)^{18-\text{r}}+\Big(\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{\text{r}}$

$={^\text{18}}\text{C}_{\text{r}}\Big((\text{x}^\frac{1}{3})\Big)^{18-\text{r}}\times\Big(\frac{1}{2}\Big)^{\text{r}}\times\Big(\frac{1}{\text{x}\frac{1}{3}}\Big)^{\text{r}}$

$={^\text{18}}\text{C}_{\text{r}}(\text{x})^{\frac{18-\text{r}}{3}}\times\Big(\frac{1}{2}\Big)^{\text{r}}$

$={^\text{18}}\text{C}_{\text{r}}(\text{x})^{\frac{18-\text{r}}{3}}\times\Big(\frac{1}{2}\Big)^{\text{r}}$

If it is independent of x, we must have

$\Rightarrow \frac{18-2\text{r}}{3}=0$

$\Rightarrow 18=2\text{r}$

$\Rightarrow \text{r}=9$

Terem independent of x = T₉₊₁ = T₁₀

Now,

$\text{T}_{10}={^\text{18}}\text{C}_{\text{9}}\big(\sqrt[3]{\text{x}}\big)^{18-9}\Big(\frac{1}{2\sqrt[3]{\text{x}}}\Big)^{9}$

$={^\text{18}}\text{C}_{\text{9}}\big(\sqrt[3]{\text{x}}\big)^{9}\times\frac{1}{2^{9}}\times\Big(\frac{1}{\sqrt[3]{\text{x}}}\Big)^{9}$

$=\frac{{^\text{18}}\text{C}_{\text{9}}}{2^{9}}$

In the binomial expansion of $\Big(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}},\big[(\text{n}+1)-7+1\big]^\text{th}\text{i.e.,}\ (\text{n}-5)^{\text{th}}$ term from the beginning is the 7^th term from the end.

Now,

$\text{T}_{7}={^\text{n}}\text{C}_{\text{6}}\Big(\sqrt[3]{2}\Big)^{\text{n}-6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{6}={^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}$

And,

$\text{T}^{\text{n}-5}={^\text{n}}\text{C}_{\text{n}-6}\Big(\sqrt[3]{2}\Big)^{6}\Big(\frac{1}{\sqrt[3]{3}}\Big)^{\text{n}-6}={^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}$

It is given that,

$\frac{\text{T}^{7}}{\text{T}^{\text{n}-5}}=\frac{1}{6}$

$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{6}}\times2^{\frac{\text{n}}{3}-2}\times\frac{1}{3^{2}}}{{^\text{n}}\text{C}_{\text{6}}\times2^{2}\times\frac{1}{3^{\frac{\text{n}}{3}-2}}}=\frac{1}{6}$

$\Rightarrow 2^{\frac{\text{n}}{3}-2-2}\times3^{\frac{\text{n}}{3}-2-2}=\frac{1}{6}$

$\Rightarrow \Big(\frac{1}{6}\Big)^{4-\frac{\text{n}}{3}}=\frac{1}{6}$

$\Rightarrow 4-\frac{\text{n}}{3}=1$

$\Rightarrow \text{n}=9$

Hence, the vaue of n is 9.

Suppose the (r + 1)th term in the given expression in independent of x.

Now,

$\Big(2\text{x}+\frac{1}{3\text{x}^{2}}\Big)^{9}$

$\text{T}_{\text{r+1}}= {^\text{9}}\text{C}_{\text{r}}\big(2\text{x})^{9-\text{r}}\Big(\frac{1}{3\text{x}^{2}}\Big)^{\text{r}}$

$={^\text{9}}\text{C}_{\text{r}}.\frac{2^{9-\text{r}}}{3^{\text{r}}}\ \text{x}^{\text{9-r}}\Big(\frac{1}{3\text{x}^{2}}\Big)^{\text{r}}$

For ths term to be independent of x, we must have

9 - 36 = 0

⇒ r = 3

Hence, the required term is the 4th term.

Now, we have

${^\text{9}}\text{C}_{\text{3}} \frac{2^{6}}{3^{3}}$

$={^\text{9}}\text{C}_{\text{3}}\times\frac{64}{27}$

For the given binomial expansion = 9.

So middle term are $\Big(\frac{9+1}{2}\Big)=5^{\text{th}}$ and $\Big(\frac{9+3}{2}\Big)=6^{\text{th}}$

$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{\text{x}}\Big)^{9-4}\Big(\frac{\text{x}}{\text{p}}\Big)^{4}$

$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{\text{x}}\Big)^{5}\Big(\frac{\text{x}}{\text{p}}\Big)^{4}$

$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{\text{x}}\Big)$

$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{5}}\Big(\frac{\text{p}}{\text{x}}\Big)^{9-5}\Big(\frac{\text{x}}{\text{p}}\Big)^{5}$

$\text{T}_{\text{5}}= {^\text{9}}\text{C}_{\text{5}}\Big(\frac{\text{p}}{\text{x}}\Big)^{4}\Big(\frac{\text{x}}{\text{p}}\Big)^{5}$

$\text{T}_{\text{6}}= {^\text{9}}\text{C}_{\text{5}}\Big(\frac{\text{x}}{\text{p}}\Big)$

$\text{T}_{\text{6}}=\frac{126}{\text{x}}$

The middle terms are $\frac{126}{\text{x}}.$

$\Big(\text{x}+\sqrt{\text{x}^2-1}\Big)^6+\Big(\text{x}-\sqrt{\text{x}^2-1}\Big)^6$

$=2\Big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_2\text{x}^4\big(\sqrt{\text{x}^2-1}\big)^2+{^6\text{C}}_4\text{x}^2\big(\sqrt{\text{x}^2-1}\big)^4+{^6\text{C}}_6\big(\sqrt{\text{x}^2-1}\big)^6\Big]$

$=2\Big[\text{x}^6+15\text{x}^4\big(\text{x}^2-1\big)+15\text{x}^2\big(\text{x}^2-1\big)^2+\big(\text{x}^2-1\big)^3\Big]$

$=2\Big[\text{x}^6+15\text{x}^6-15\text{x}^4+15\text{x}^6+15\text{x}^2-30\text{x}^4+\text{x}^6-1-3\text{x}^4+3\text{x}^2\Big]$

$=64\text{x}^6-96\text{x}^4+36\text{x}^2-2$

We have,

$(101)^4=(100+1)^4$

$={^4\text{C}}_0\times100^4+{^4\text{C}}_1\times100^3+{^4\text{C}}_2\times100^2+{^4\text{C}}_3\times100+{^4\text{C}}_4$

$=100^4+4\times100^3+6\times100^2+4\times100+1$

$=100000000+4000000+60000+400+1$

$=104060401$

$\therefore(101)^4=104060401$

$3^{2\text{n}+2}-8\text{n}-9$

$=3^{2(\text{n}+1)}-8\text{n}-9$

$=9^{\text{n}+1}-8\text{n}-9$

$=(1+8)^{\text{n}+1}-8\text{n}-9$

$\Big({^{\text{n}+1}}\text{C}_0+{^{\text{n}+1}}\text{C}_18^1+{^{\text{n}+1}}\text{C}_28^2+.....+{^{\text{n}+1}}\text{C}_{\text{n}+1}8^{\text{n}+1}\Big)-8\text{n}-9$

$\Big(1+8(\text{n}+1)64^{\text{n}+1}\text{C}_2+64(8)^{\text{n}-1}\Big)-8\text{n}-9$

$=64\Big({^{\text{n}+1}\text{C}}_2+......+8^{\text{n}-1}\Big)$

Thus, $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64.

$\Big(2\text{ax}-\frac{\text{b}}{\text{x}^{2}}\Big)^{12}$

For the given term is $\Big(\frac{12}{2}+1\Big)=7$

$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}(2\text{ax})^{12-6}\ \Big(\frac{\text{b}}{\text{x}^{2}}\Big)^{6}$

$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}(2\text{ax})^{6}\ \Big(\frac{\text{b}}{\text{x}^{2}}\Big)^{6}$

$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}(2^{6}\text{a}^{6}{\text{x}}^{6})\ \Big(\frac{\text{b}}{\text{x}^{12}}\Big)^{6}$

$\text{T}_{\text{7}}= {^\text{12}}\text{C}_{\text{6}}\Big(\frac{2^{6}\text{a}^{6}\text{b}^{6}}{\text{x}^{6}}\Big)$

Let (r + 1)^th term, in the expansion of $\Big(\sqrt{\text{x}}-\frac{\text{k}}{\text{x}^{2}}\Big)^{10},$ be free from x and be equal to T_r+1.

Then,

$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}(\sqrt{\text{x}})^{10-\text{r}}\Big(\frac{-\text{k}}{\text{x}^{\text{2}}}\Big)^{\text{r}}={^\text{10}}\text{C}_{\text{r}}\text{x}^{5-\frac{5\text{r}}{2}}(-\text{k})^{\text{r}}\ ...(\text{i}) $

If T_r+1 is independent of x, then

$5-\frac{5\text{r}}{2}=0$

$\Rightarrow \text{r}=2$

Putting r = 2 in (i), we obtain

$\text{T}_{3}={^\text{10}}\text{C}_{\text{2}}(-\text{k})^{2}=45\text{k}^{2}$

But it is given that the value of the term free x is 405.

$\therefore 45\text{k}^{2}=405$

$\Rightarrow \text{k}^{2}=9$

 $\Rightarrow \text{k}^{2}=\pm3$

Hence, the value of k is $\pm3.$

$\Big(\sqrt{\text{x}+1}+\sqrt{\text{x}-1}\Big)^6+\Big(\sqrt{\text{x}+1}-\sqrt{\text{x}-1}\Big)^6$

$={^6\text{C}}_0\big(\sqrt{\text{x}+1}\big)^6+{^6\text{C}}_1\big(\sqrt{\text{x}+1}\big)^5\big(\sqrt{\text{x}-1}\big)+{^6\text{C}}_2\big(\sqrt{\text{x}+1}\big)^4\\\big(\sqrt{\text{x}-1}\big)^2-{^6\text{C}}_3\big(\sqrt{\text{x}+1}\big)^3\big(\sqrt{\text{x}-1}\big)^3\\+{^6\text{C}}_4\big(\sqrt{\text{x}+1}\big)^2\big(\sqrt{\text{x}-4}\big)^4+{^6\text{C}_5\big(\sqrt{\text{x}+1}}\big)\big(\sqrt{\text{x}-1}\big)^5+{^6\text{C}}_6\big(\sqrt{\text{x}-1}\big)^6+{^6\text{C}}_0\big(\sqrt{\text{x}+1}\big)^6\\{^6\text{C}}_1\big(\sqrt{\text{x}+1}\big)^5\big(\sqrt{\text{x}-1}\big)+{^6\text{C}}_2\big(\sqrt{\text{x}+1}\big)^4\times\big(\sqrt{\text{x}-1}\big)^2-{^6\text{C}}_3\big(\sqrt{\text{x}+1}\big)^3\\\big(\sqrt{\text{x}-1}\big)63+{^6\text{C}}_4\big(\sqrt{\text{x}+1}\big)^2\big(\sqrt{\text{x}-1}\big)^4-{^6\text{C}}_5\big(\sqrt{\text{x}+1}\big)\big(\sqrt{\text{x}-1}\big){^6\text{C}}_6\big(\sqrt{\text{x}-1}\big)^6$

$=2\big[(\text{x}+1)^3+15(\text{x}+1)^2(\text{x}-1)+15(\text{x}+1)(\text{x}-1)^2+(\text{x}-1)^3\big]$

$=2\Big[\text{x}^3+1+3\text{x}+3\text{x}^2+15\text{x}^3-15\text{x}^2+15\text{x}-15+30\text{x}^2-30\text{x}\\+15\text{x}^3+15\text{x}^2+15\text{x}+15-30\text{x}^2-30\text{x}+\text{x}^3-1-3\text{x}^2+3\text{x}\Big]$

$=64\text{x}^3-48\text{x}$

$=16\text{x}(4\text{x}^2-3)$

Coefficient of (r + 1)th term in the expansion of $(4+\text{x})^{\text{n+1}}={^\text{n+1}}\text{C}_{\text{r}+1-1}={^\text{n+1}}\text{C}_{\text{r}}$ 

Coefficient of rth term in $(1+\text{x})^{\text{n}}$ + Coefficient of (r + 1)th term in $(1+\text{x})^{\text{n}}$

$={^\text{n}}\text{C}_{\text{r+1}}+{^\text{n}}\text{C}_{\text{r}-1-1}$

$={^\text{n}}\text{C}_{\text{r-1}}+{^\text{n}}\text{C}_{\text{r}}$

$=\frac{\text{n}!}{(\text{n}-(\text{r}-1)!(\text{r}-1)!}+\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}$

$=\frac{\text{n}!}{(\text{n}-\text{r}+1)!(\text{r}-1)}+\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}$

$=\frac{\text{n}!}{(\text{n}-\text{r}+1)(\text{n}-\text{r})!(\text{r}-1)!}+\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}(\text{r}-1)!}$

$=\frac{\text{n}!}{(\text{n}-\text{r}+1)(\text{n}-\text{r})!(\text{r}-1)!}+\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!\text{r}}$

$=\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!}\Big[\frac{1}{\text{n}-\text{r}+1}+\frac{1}{\text{r}}\Big]$

$=\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!}\Big[\frac{\text{r}+\text{n}-\text{r}+1}{(\text{n}-\text{r}+1)\text{r}}\Big]$

$=\frac{\text{n}!}{(\text{n}-\text{r})!(\text{r}-1)!}\Big[\frac{\text{n+1}}{(\text{n}-\text{r}+1)\text{r}}\Big]$

$=\frac{\text{n}!(\text{n}+1)}{(\text{n}-\text{r})!(\text{n}-\text{r}+1)(\text{r}-1)!\text{r}}$

$=\frac{(\text{n}+1)}{(\text{n}-\text{r}+1)!\text{r}!}$

$=\frac{(\text{n}+1)}{(\text{n}+1-\text{r})!\text{r}!}$

$={^\text{n+1}}\text{C}_{\text{r}}$

${^\text{n+1}}\text{C}_{\text{r}}={^\text{n}}\text{C}_{\text{r}-1}+{^\text{n}}\text{C}_{\text{r}}$

$2^{4\text{n}+4}- 15\text{n}-16=2^{4(\text{n}+1)}-15\text{n}-15-1$

$=(16)^{\text{n}+1}-15(\text{n}+1)-1$

$=(1+15)^{\text{n}+1}-15(\text{n}+1)-1$

$=\big[{^{\text{n}+1}\text{C}}_0+{^{\text{n}+1}\text{C}}_1(15)+{^{\text{n}+1}\text{C}}_2(15)^2+.....\\+{^{\text{n}+1}\text{C}}_{\text{n}+1}(15)^{\text{n}+1}\big]-15(\text{n}+1)-1$

$=\big[1+15({\text{n}+1})+{^{\text{n}+1}\text{C}}_2(15)^2+.....\\+{^{\text{n}+1}\text{C}}_{\text{n}+1}(15)^{\text{n}+1}\big]-15(\text{n}+1)-1$

$=225\big[{^{\text{n}+1}\text{C}}_2+.....{^{\text{n}+1}\text{C}}_{\text{n}+1}(15)^{\text{n}-1}\big]$

= 225 × natural number

In the binomial expansion of $\Big(\frac{\text{p}}{2}+2\Big)^{8},$ we observe that $\Big(\frac{8}{2}+1\Big)$ i.e., 5th term is the middle term.

It is given that the middle term is 1120.

$\therefore \text{T}_{5}=1120$

$\Rightarrow \therefore\ {^\text{8}}\text{C}_{\text{4}}\Big(\frac{\text{p}}{2}\Big)^{8-4}(2)^{4}=1120$

$\Rightarrow \text{p}^{4}=16$

$\Rightarrow \text{p}=\pm2$

Hence, the real value of p is $\pm2.$

$\text{T}_{\text{r+1}}= {^\text{9}}\text{C}_{\text{r}}\Big(\frac{3\text{x}^{2}}{2}\Big)^{9-\text{r}}\Big(\frac{-1}{3\text{x}}\Big)^{\text{r}}$

$= {^\text{9}}\text{C}_{\text{r}}\Big(\frac{3}{2}\Big)^{9-\text{r}}\big(\text{x}^{18-2\text{r}}\big)\Big(\frac{-1}{3}\Big)^{\text{r}}\text{x}^{-\text{r}}$

Let $\text{T}_{\text{r+1}}$ be independent of x.

18 - 3r = 0 or r = 6

$\therefore$ Required term,

 $\text{T}_{\text{r+1}}=\text{T}_{\text{6+1}}=\text{T}_{\text{7}}= {^\text{9}}\text{C}_{\text{6}}\Big(\frac{3}{2}\Big)^{9-\text{6}}\Big(\frac{-1}{3}\Big)^{\text{6}}\text{x}^{18-3(6)}$

$=84\Big(\frac{27}{8}\Big)\Big(\frac{1}{179}\Big)\text{x}^{0}=\frac{7}{18}$

We have,

$\Big(\text{x}+\frac{1}{\text{x}}\Big)^{2\text{n}}$

Let $(\text{r}+1)^{\text{2n}}$ term be independent of x.

$\text{T}_{\text{r}+1}={^\text{2n}\text{C}}_\text{r}(\text{x})^{\text{2n}-\text{r}}\Big(\frac{1}{\text{x}}\Big)^{\text{r}}$

$={^\text{2n}\text{C}}_\text{r}(\text{x})^{\text{2n}-\text{r}-\text{r}}$

$={^\text{2n}\text{C}}_\text{r}(\text{x})^{\text{2n}-\text{2r}}$

If it is independent of x, we must have,

$2\text{n}-2\text{r}=0$

$\Rightarrow 2\text{n}=2\text{r}$

$\Rightarrow \text{r}=\text{n}$

Term independent of $\text{x}=\text{T}_{\text{n+1}}$

$={^\text{2n}\text{C}}_\text{n}$

$=\frac{(2\text{n})!}{(2\text{n}-\text{n})!\text{n}!}$

$=\frac{(2\text{n})!}{\text{n}!\text{n}!}$

$=\frac{(2\text{n})(2\text{n}-1)(2\text{n}-2)....5\times4\times3\times2\times1}{\text{n}!\text{n}!}$

$=\frac{1\times3\times5\times.....(2\text{n}-1)(2\times4\times6\times...2\text{n})}{\text{n!}\text{n}!}$

$=\frac{1\times3\times5\times.....(2\text{n}-1)\ \times\ 2^{\text{n}}(1\times2\times3\times...\text{n})}{\text{n!}\text{n}!}$

$=2^{\text{n}}\times\frac{(1\times3\times5\times....(2\text{n}-1))}{\text{n}!}$

The Term independent to $\text{x}=2^{\text{n}}\times\frac{(1\times3\times5\times....(2\text{n}-1))}{\text{n}!}$ 

Hence proved.

$\Big(\text{x}-\frac{1}{\text{x}}\Big)^{2\text{n}+1}$

2n + 1 is odd hence this expansion will have 2n + 2 = even terms.

Hence, middle terms is $\frac{2\text{n}+1}{2}=\text{n+1},\text{n+2}$

Term formula is,

$\text{T}_{\text{n}}=\text{T}_{\text{r}+1}=(-1)^{r}\ {^\text{n}}\text{C}_{\text{r}}\text{x}^{\text{n-r}}\ \text{y}^{\text{r}}$

$\text{T}_{\text{n+1}}=\text{T}_{\text{n}+1}=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n}}\text{x}^{\text{2n+1}}\ \Big(\frac{1}{\text{x}}\Big)^{\text{n}}$

$=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n}}\ \text{x}^{\text{n+1-n}}$

$=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n}}\text{x}$

$\text{T}_{\text{n+2}}=\text{T}_{\text{n}+1+1}=(-1)^{\text{n}}\ {^\text{2n+1}}\text{C}_{\text{n+1}}\text{x}^{\text{2n+1}-\text{n-1}}\ \Big(\frac{1}{\text{x}}\Big)^{\text{n+1}}$

$=(-1)^{\text{n+1}}\ {^\text{2n+1}}\text{C}_{\text{n+1}}\frac{1}{\text{x}}$

$=(-1)^{\text{n+1}}\ {^\text{2n+1}}\text{C}_{\text{n}}\frac{1}{\text{x}}$

$(\sqrt3+\sqrt2)^6-(\sqrt3-\sqrt2)^6$

$=2\big[{^6\text{C}}_1(\sqrt3)^5(\sqrt2)+{^6\text{C}}_3(\sqrt3)^3(\sqrt2)^3+{^6\text{C}}_5(\sqrt3)(\sqrt2)^5\big]$

$=2\big[6\times\sqrt6\times9+20\times3\sqrt3\times2\sqrt2+6\times\sqrt3\times4\sqrt2\big]$

$=2\big[54\sqrt6+120\sqrt6+24\sqrt6\big]$

$=2\big[198\sqrt6\big]$

$=396\sqrt6$

$(\text{a}+\text{b})^4-(\text{a}-\text{b})^4$

$=\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2{\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]$

$=\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]\\-\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})^2+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]$

$=\big[{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4-{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4$

$={^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\\-{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}-{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4\big]$

$=2\big[{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_3\text{ab}^3\big]$

$=2\big[4\text{a}^3\text{b}+4\text{ab}^3\big]$

$=8\big[\text{a}^3\text{b}+\text{ab}^3\big]$

$\therefore(\text{a+b})^4-(\text{a-b})^4=8(\text{a}^3\text{b}+\text{ab}^3)$

Putting $\text{a}=\sqrt3$ and $\text{b}=\sqrt2$ in equation (i) we get

$(\sqrt3+\sqrt2)^4-(\sqrt3-\sqrt2)^4=8\Big[(\sqrt3)^3\times\sqrt2+(\sqrt3)\times(\sqrt2)^2\Big]$

$=8\Big[3\sqrt6+2\sqrt6\Big]$

$=8\times5\sqrt6$

$=40\sqrt6$

$\therefore(\sqrt{3}+\sqrt2)^4-(\sqrt{3}-\sqrt2)^440\sqrt6.$

$\text{T}_{\text{r+1}}=(-1)^{\text{r}} \ {^\text{n}}\text{C}_{\text{r}}\big(2\text{x}^{2})^{25-\text{r}}\Big(\frac{3}{\text{x}^{2}}\Big)^{\text{r}}$

$=(-1)^{\text{r}} \ {^\text{n}}\text{C}_{\text{r}}\big(2)^{25-\text{r}}\ 3^{\text{r}}\ \text{x}^{50-2\text{r}-3\text{r}}$

Term independent of x = x⁰

$\Rightarrow \text{x}^{50-50}=\text{x}^{0}$

$\Rightarrow 50-5\text{r}=0$

$\Rightarrow \text{r}=10$

$\text{T}_{11}=(-1)^{\text{10}} \ {^\text{25}}\text{C}_{\text{10}}\big(2)^{15}\times 3^{\text{10}}$

$={^\text{25}}\text{C}_{\text{10}}\big(2)^{15}\times 3^{\text{10}}$

We have,

T₁ = 729, T₂ = 7290 and T₃ = 30375

According to the quation, 

 ${^\text{n}}\text{C}_{\text{0}}\text{a}^{\text{n}}\text{b}^{0}=729$

$\Rightarrow \text{a}^{\text{n}}=729$

$\Rightarrow \text{a}^{\text{n}}=3^{6}$

${^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\text{b}^{1}=7290$

${^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}=30375$

Also,

$\Rightarrow \frac{{^\text{n}}\text{C}_{\text{2}}\text{x}^{\text{n}-2}\text{a}^{2}}{{^\text{n}}\text{C}_{\text{1}}\text{x}^{\text{n}-1}\text{a}^{1}}=\frac{30375}{7290}$

$\Rightarrow \frac{\text{n}-1}{2}\times\frac{\text{b}}{\text{a}}=\frac{25}{6}\ ...(\text{i})$

$\Rightarrow \frac{\big(\text{n}-1\big)\text{b}}{\text{a}}=\frac{25}{3}$

And,

$\frac{{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}\text{b}^{1}}{{^\text{n}}\text{C}_{\text{0}}\text{a}^{\text{n}}\text{b}^{0}}=\frac{7290}{729}$

$\Rightarrow \frac{\text{n}\text{b}}{\text{a}}=\frac{10}{1}\ ...(\text{ii})$

On dividing (ii) by (i), we get

$\frac{\frac{\text{nd}}{\text{a}}}{\frac{(\text{n}-1)\text{b}}{\text{a}}}=\frac{10\times3}{25}$

$\Rightarrow \frac{\text{n}}{\text{n}-1}=\frac{6}{5}$

$\Rightarrow \text{n}=6$

Since, $\text{a}^{6}=3^{6}$

Hence, $\text{a}=3$

Now,

$\frac{\text{nb}}{\text{a}}=10$

$\Rightarrow \text{b}=5$

$(3+\sqrt2)^5-(3-\sqrt2)^5$

$=2\big[{^5\text{C}}_1(3)^4(\sqrt2)^1+{^5\text{C}}_3(3)^2(\sqrt2)^3+{^5\text{C}}_5(\sqrt3)^5\big]$

$=2\big[5\times81\times\sqrt2+10\times9\times2\sqrt2+4\sqrt2\big]$

$=2\big[405\sqrt2+180\sqrt2+4\sqrt2\big]$

$=2\big[589\sqrt2\big]$

$=1178\sqrt2$

For the given binomial expansion = 10.

So middle term are $\Big(\frac{10}{2}+1\Big)=6^{\text{th}}$

$\text{T}_{\text{5}}= {^\text{10}}\text{C}_{\text{5}}\Big(\frac{\text{p}}{\text{x}}\Big)^{10-5}\Big(-\frac{\text{a}}{\text{x}}\Big)^{5}$

$\text{T}_{\text{6}}= -{^\text{10}}\text{C}_{\text{5}}\Big(\frac{\text{x}}{\text{a}}\Big)^{5}\Big(\frac{\text{a}}{\text{x}}\Big)^{5}$

$\text{T}_{\text{5}}= {^\text{10}}\text{C}_{\text{5}}=-252$

The middle terms are -252.

We have,

$(1+\text{x})^{\text{2n}}$

Now,

Coefficient of 2nd term $={^\text{2n}}\text{C}_{\text{2}-1}={^\text{2n}}\text{C}_{\text{1}}$

Coefficient of 3rd term $={^\text{2n}}\text{C}_{\text{3}-1}={^\text{2n}}\text{C}_{\text{2}}$

Coefficient of 4th term $={^\text{2n}}\text{C}_{\text{4}-1}={^\text{2n}}\text{C}_{\text{3}}$

It is given that these coefficients are in A.P.

$\therefore\ 2\ {^\text{2n}}\text{C}_{\text{2}}={^\text{2n}}\text{C}_{\text{1}}+{^\text{2n}}\text{C}_{\text{3}}$

$\Rightarrow 2=\frac{{^\text{2n}}\text{C}_{\text{1}}}{{^\text{2n}}\text{C}_{\text{2}}}+\frac{{^\text{2n}}\text{C}_{\text{3}}}{{^\text{2n}}\text{C}_{\text{2}}}$

$\Rightarrow\ 2=\frac{2}{2\text{n}-2+1}+\frac{2\text{n}-3+1}{3}$

$\Rightarrow 2=\frac{6+(2\text{n}+1)(2\text{n}-2)}{3(2\text{n}-1)}$

$\Rightarrow 6(2\text{n}-1)=6+4\text{n}^{2}-4\text{n}-2\text{n}+2$

$\Rightarrow 12\text{n}-6=8+4\text{n}^{2}-6\text{n}$

$\Rightarrow 4\text{n}^{2}-6\text{n}-12\text{n}+8+6=0$

$\Rightarrow 4\text{n}^{2}-18\text{n}+14=0$

$\Rightarrow 2(2\text{n}^{2}-9\text{n}+7)=0$

$\Rightarrow 2\text{n}^{2}-9\text{n}+7=0$

Hence proved.

We know that the coefficient of rth term in the expansion of $(1+\text{x}){^\text{n}}$ is ${^\text{n}}\text{C}_{\text{r}-1}$

Coefficient of (2r + 4)th term of the expansion $(1+\text{x})^{18}={^\text{18}}\text{C}_{\text{2r}+4-1}={^\text{18}}\text{C}_{\text{2r}+3}$

Coefficient of (r - 2)th term of the expansion $(1+\text{x})^{18}={^\text{18}}\text{C}_{\text{r}-2-1}={^\text{18}}\text{C}_{\text{r}-3}$

It is given that these coefficients are equal.

$\Rightarrow {^\text{18}}\text{C}_{\text{2r}+3}={^\text{18}}\text{C}_{\text{r}-3}$

$\Rightarrow 2\text{r}+3=\text{r}-3$ or $2\text{r}+3+\text{r}-3=18$

$\Rightarrow \text{r}=-6$ or $\text{r}=6$

$\Rightarrow\text{r}=6$

$(0.99)^5+(1.01)^5$

$=(1-.01)^5+(1+.01)^5$

$=2\big[{^5\text{C}}_1+{^5\text{C}}_3(.01)^2+{^5\text{C}}_5(.01)^5\big]$

$=2\Big[5+10\times\frac{1}{10^4}+\frac{1}{10^{10}}\Big]$

$=2\Big[5+\frac{1}{1000}+\frac{1}{10^{10}}\Big]$

$=2.0020001$

We have,

$(\text{x}+1)^6+(\text{x}-1)^6$

$=\Big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4+{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2+{^6\text{C}}_5\text{x}^1+{^6\text{C}}_6\text{X}^0\big]\\+\big[{^6\text{C}}_0\text{x}^6(-1)^0+{^6\text{C}}_1\text{x}^5(-1)^1+{^6\text{C}}_2\text{x}^4(-1)^2+{^6\text{C}}_3\text{x}^3(-1)^3\\+{^6\text{C}}_4\text{x}^2(-1)^4+{^6\text{C}}_5\text{x}^1(-1)^5+{^6\text{C}}_6\text{x}^0(-1)^6\big]\\{^6\text{C}}_0\text{x}^6-{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4-{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2-{^6\text{C}}_5\text{x}+{^\text{C}}_6\Big]$

$=\Big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4+{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2+{^6\text{C}}_5\text{x}+{^6\text{C}}_6\\{^6\text{C}}_0\text{x}^6-{^6\text{C}}_1\text{x}^5+{^6\text{C}}_2\text{x}^4-{^6\text{C}}_3\text{x}^3+{^6\text{C}}_4\text{x}^2-{^6\text{C}}_5\text{x}+{^6\text{C}}_6\Big]$

$=2\big[{^6\text{C}}_0\text{x}^6+{^6\text{C}}_2\text{x}^4+{^6\text{C}}_4\text{x}^2+{^6\text{C}}_6\big]$

$=2\big[\text{x}^6+15\text{x}^4+15\text{x}^2+1\big]$

$\therefore(\text{x}+1)^6+(\text{x}-1)^6=2\big[\text{x}^6+15\text{x}^4+15\text{x}^2+1\big]...(\text{i})$

Putting $\text{x}=\sqrt2$ in equation (i), we get

$(\text{x}+1)^6+(\text{x}-1)^6=2\big[(\sqrt2)^6+15(\sqrt2)^4+15(\sqrt2)^2+1\big]$

$=2\big[8+60+30+1\big]$

$=2\big[99\big]$

$=198$

$\therefore(\text{x}+1)^6+(\text{x}-1)^6=198$

We have,

$(1.1)^{10000}=(1+0.1)^{10000}$

$={^{10000}\text{C}}_0+{^{10000}\text{C}}_1(0.1)+{^{10000}\text{C}}_2(0.1)^2+...+{^{10000}\text{C}}_{10000}(0.1)^{10000}$

= 1 + 10000 × (0.1) + other positive terms

= 1 + 1000 + other positive terms

= 1001 + other positive terms > 1000

$\therefore$ (1.1)¹⁰⁰⁰⁰ > 1000

$(1.2)^{4000}=(1+0.2)^{4000}$

$={^{4000}\text{C}}_0(0.2)^0(1)^{4000}+{^{4000}\text{C}}_1(0.2)^1{1}^{3999}+....\\+{^{4000}\text{C}}_400(0.2)^0(1)^{4000}(0.2)^{4000}1^0$

$=1 + 4000 \times 0.2 \times 1 +.......+(0.2)^{4000}$

$= 1 + 800 +.......+(0.2)^{4000}$

Here, we dearty observe (1, 2)⁴⁰⁰⁰ is less than (801) thus, (1, 2)⁴⁰⁰⁰ < 800.

We have,

$(96)^3=(100-4)^3$

$={^3\text{C}}_0\times100^3+{^3\text{C}}_1\times100^2\times(-4)+{^3\text{C}}_2\times100\times(-4)^2+{^3\text{C}}_3\times(-4)^3$

$=100^3-3\times100^2\times4+3\times100\times4^2-4^3$

$=1000000-120000+4800-64$

$1004800-120064$

$=884736$

$\therefore(96)^3=884736$