MATHS 3 Marks Question

The expansion of (x + y)ⁿ has n + 1 term so, the expansion of $\Big(\text{x}-\frac{1}{\text{x}}\Big)^6$ has 7 terms.

Using binomial theorem, we have

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

$\text{x}^6-6\text{x}^4+15\text{x}^2-20+\frac{15}{\text{x}^2}-\frac{6}{\text{x}^4}+\frac{1}{\text{x}^6}$

$(1-3\text{x}+7\text{x}^2)(1+\text{x})^{16}=(1-3\text{x}+7\text{x}^2)\${^{16}\text{C}}_0-{^{16}\text{C}}_1\text{x}+{^\text{n}\text{C}}_2\text{x}^2+.....+{^{16}\text{C}}_{16}\text{x}^{16})$

$\therefore\ \text{Coefficient}\ \text{of}\ \text{x}\ \text{in}\ (1-3\text{x}+7\text{x}^2)(1-\text{x})^{16}$

$=1\times(-{^{16}\text{C}}_1)-3\times(-{^{16}\text{C}}_0)$

$=-16-3$

$=-19$

Suppose $\text{r}^{\text{th}}, (\text{r}+1)^{\text{th}}$ and $(\text{r}+2)^{\text{th}}$ terms are the three consecutive terms.

Their respective coefficients are $​{^\text{n}}\text{C}_{\text{r}-1}, ​{^\text{n}}\text{C}_{\text{r}}$ and $​{^\text{n}}\text{C}_{\text{r}+1}.$

We have,

$​{^\text{n}}\text{C}_{\text{r}-1}= ​{^\text{n}}\text{C}_{\text{r}+1}=56$

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

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

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

Now,

$​{^\text{n}}\text{C}_{\frac{\text{n}}{2}}=70$ and $​{^\text{n}}\text{C}_{\Big(\frac{\text{n}}{2}-1\Big)}=56$

$\Rightarrow \frac{​{^\text{n}}\text{C}_{\frac{\text{n}}{2}-1}}{​{^\text{n}}\text{C}_{\frac{\text{n}}{2}}}=\frac{56}{70}$

$\Rightarrow \frac{\frac{\text{n}}{2}}{\Big(\frac{\text{n}}{2}+1\Big)}=\frac{8}{10}$

$\Rightarrow 5\text{n}=4\text{n}+8$

$\Rightarrow \text{n}=8$

So, $\text{r}=\frac{\text{n}}{2}=4$

Thus, the required term 4th, 5th and 6th.

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 $\frac{126}{\text{x}}.$

$(2+\sqrt3)^7+(2-\sqrt3)^7$

$=2\Big[{^7\text{C}}_02^7+{^7\text{C}}_2{2}^5(\sqrt3)^2+{^7\text{C}}_4(2)^4(\sqrt3)^4+{^7\text{C}}_6{2}(\sqrt{3})^6\Big]$

$=2\Big[128+21\times32\times3+35\times8\times9+7\times2\times27\Big]$

$=2\big[128+2016+2520+378\big]$

$=2\big[5042\big]$

$=10084$

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

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

Indeprndent of x = x⁰

$\text{x}^{3\text{n}-3\text{r}}=\text{x}^{0}$

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

$=(-1)^{\text{n}}\ {^\text{3n}}\text{C}_{\text{r}}$

$(1+3\text{x}+3\text{x}^{2}+\text{x}^{3})^{2\text{n}}$

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

Here, n is an even number.

Middle term $=(\frac{6\text{n}}{2}+1\big)=(3\text{n}+1)$

Now, we have,

$\text{T}_{3\text{n}+1}$

$={^\text{6n}}\text{C}_{3\text{n}}\text{x}^{3\text{n}}$

$=\frac{6\text{n}!}{3\text{n}!}\text{x}^{3\text{n}}$

$\Big(\frac{\text{x}}{3}+9\text{y}\Big)^{10}$

Hence n = 10, which is even therefore it has 11 terms.

Middle term is $\Big(\frac{\text{n}}{2}+1\Big)=6^\text{k}$

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

$\text{T}_{\text{6}}=\text{T}_{\text{5}+1}=(-1)^{\text{5}}\ {^\text{10}}\text{C}_{\text{5}}\Big(\frac{\text{x}}{3}\Big)^{10-5}(9\text{y})^{5}$

$=-\frac{10!}{5!5!}\times\frac{\text{x}^{5}}{3^{5}}\times9^{5}\times\text{y}^{5}$

$=61236\text{x}^{5}\text{y}^{5}$

 $\Big(3{\text{x}}-\frac{\text{x}^3}{6}\Big)^9$

Here, n = 9 which odd number

$\therefore\Big(\frac{9+1}{2}\Big)^\text{th}\ \text{and}\ \Big(\frac{9+1}{2}+1\Big)^\text{th}$ i.e., 5^th, 6^th term are the middle term.

Here, the term formula is

$\text{T}_5=\text{T}_{4+1}=({-1})^4\ {^9\text{C}}_4(3\text{x})^5\Big(\frac{\text{x}^3}{6}\Big)^4$

$={^9\text{C}}_4\frac{3^5}{6^4}\times\text{x}^5\times\text{x}^{12}$

$=\frac{9\times8\times7\times6\times3^5}{4\times3\times2\times3^4\times2^4}\text{x}^{17}$

$=\frac{189}{8}\text{x}^{17}$

$\text{T}_6=\text{T}_{5+1}=(-1)^5\ {^9\text{C}}_5(3\text{x})^4\Big(\frac{\text{x}^3}{6}\Big)^5$

$=-\frac{9\times8\times7\times6}{5\times4\times3\times2}\times\frac{3^4}{6^5}\times\text{x}^4\times\times^{15}$

$=-\frac{9\times8\times7\times6\times3^4}{5\times4\times3\times2\times3^5\times2^5}\text{x}^{19}$

$=\frac{-21}{16}\text{x}^{19}$ 

Let y = 1 + 2x, then

$(1+2\text{x}-3\text{x}^2)^5=(\text{y}-3\text{x}^2)^5$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of (y - 3x²)⁵ has 6 terms.

Using binomial theorem to expand, we get

$(\text{y}-3\text{x}^2)^5={^5\text{C}}_0\text{y}^5(3\text{x}^2)^0-{^5\text{C}}_1\text{y}^4(3\text{x}^2)^1+{^5\text{C}}_2\text{y}^3(3\text{x}^2)^2\\-{^5\text{C}}_3\text{y}^2(3\text{x}^2)^3+{^5\text{C}}_4\text{y}(3\text{x}^2)^4-{^5\text{C}}_5\text{y}^0(3\text{x}^2)^5$

$=\text{y}^5-5\text{y}^43\text{x}^2+10\text{y}^39\text{x}^4-10\text{y}^2(27\text{x}^6)+5\text{y}81\text{x}^8-243\text{x}^{10}$

Now,

$\text{y}^5=(1+2\text{x})^5={^5\text{C}}_0+{^5\text{C}}_1(2\text{x})^1+{^5\text{C}}_2(2\text{x})^2\\+{^5\text{C}}_3(2\text{x})^3+{^5\text{C}}_4(2\text{x})^4+{^5\text{C}}_5(2\text{x})^5$

$\text{y}^4=(1+2\text{x})^4={^4\text{C}}_0+{^4\text{C}}_1(2\text{x})^1 +{^4\text{C}}_2(2\text{x})^2+{^4\text{C}}_3(2\text{x})^3+{^4\text{C}}_4(2\text{x})^4$

$\text{y}^3=(1+2\text{x})^3={^3\text{C}}_0+{^3\text{C}}_1(2\text{x}) +{^3\text{C}}_2(2\text{x})^2+{^3\text{C}}_3(2\text{x})^3$

$\text{y}^2=(1+2\text{x})^2={^2\text{C}}_0+{^2\text{C}}_1(2\text{x}) +{^2\text{C}}_2(2\text{x})^2$

$\text{y}=(1+2\text{x})$

Substituting the valus of powers of y in the equation above we get,

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

$-15\text{x}^2\big[{^4\text{C}}_0+{^4\text{C}}_1(2\text{x})^1+{^4\text{C}}_2(2\text{x})^2+{^4\text{C}}_3(2\text{x})^3+{^4\text{C}}_4(2\text{x})^4\big]\\+90\text{x}^4\big[{^3\text{C}}_0+{^3\text{C}}_1(2\text{x})+{^3\text{C}}_2(2\text{x})^2+{^3\text{C}}_3(2\text{x})^3\big]-270\text{x}^6$$\big[{^2\text{C}}_0+{^2\text{C}}_1(2\text{x})+{^2\text{C}}_2(2\text{x})^2+5\times81\text{x}^8(1+2\text{x})-243\text{x}^{10}\big]$

$=10+10\text{x}+10\times4\text{x}^2+10\times8\text{x}^3+5\times16\text{x}^4\\+32\text{x}^5-15\text{x}^2-120\text{x}^3-180\text{x}^4+480\text{x}^5-240\text{x}^6+90\text{x}^4+540\text{x}^5\\+32\text{x}^5-15\text{x}^2-120\text{x}^3-180\text{x}^4+480\text{x}^5-240\text{x}^6+90\text{x}^4+540\text{x}^5\\+1080\text{x}^6+720\text{x}^7-270\text{x}^6-1080\text{x}^7-1080\text{x}^8+405\text{x}^8+810\text{x}^9-243\text{x}^{10}$

$=1+10\text{x}+25\text{x}^2-40\text{x}^3-190\text{x}^4+92\text{x}^5+570\text{x}^6\\-360\text{x}^7-675\text{x}^8+810\text{x}^9-243\text{x}^{10}$

$(\sqrt3+1)^5-(\sqrt3-1)^5$

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

$=2\big[5\times9+10\times3+1\big]$

$=2\big[45+30+1\big]$

$=2\big[76\big]$

$=152$

$\Big({\text{x}^4}-\frac{1}{\text{x}^3}\Big)^{11}$

Here, n = 11, which is odd number

$\therefore\Big(\frac{11+1}{2}\Big)^\text{th}\ \text{and}\ \Big(\frac{11+1}{2}+1\Big)^\text{th}$ = 6^th, 7^th term are the middle terms in $\Big({\text{x}^4}-\frac{1}{\text{x}^3}\Big)^{11}$

The term formula is

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

$\text{T}_6=\text{T}{5+1}=(-1)^5\ {^{11}\text{C}}_5(\text{x}^4)^{11-5}\Big(\frac{1}{\text{x}^3}\Big)^5$

$=-{^{11}\text{C}}_5\text{x}^{24}\frac{1}{\text{x}^{15}}$

$=\frac{-11\times10\times9\times8\times7}{5\times4\times3\times2\times1}\text{x}^9$

$=(-11\times3\times2\times7)\text{x}^9$

$=(-462)\text{x}^9$

$=-462\text{x}^9$

$\text{T}_7=\text{T}_{6+1}=(-1)^6\ {^{11}\text{C}}_6(\text{x}^4)^{11-6}\Big(\frac{1}{\text{x}^3}\Big)^6$

$=462\frac{\text{x}^{20}}{\text{x}^{18}}$

$=462\text{x}^2$

 $\Big(3{\text{x}}-\frac{2}{\text{x}^2}\Big)^{15}$

7^th and 8^th terms are middle terms

$\Big(\frac{15}{7}\Big)(3\text{x})^8\Big(-\frac{2}{\text{x}^2}\Big),\Big(\frac{15}{8}\Big)(3\text{x}^7)\Big(-\frac{2}{\text{x}^2}\Big)^8$

$\frac{-6435\times3^8\times2^7}{\text{x}^6},\frac{6337\times3^7\times2^8}{\text{x}^9}$

 $\Big(2{\text{x}^2}-\frac{1}{\text{x}}\Big)^7$

Here, n = 7 which odd

$\therefore\Big(\frac{7+1}{2}\Big)^\text{th}\ \text{and}\ \Big(\frac{7+1}{2}+1\Big)^\text{th}$ i.e., 4^th, 5^th term are the middle term $\Big(2{\text{x}^2}-\frac{1}{\text{x}}\Big)^7$

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

$\text{T}_4=\text{T}_{3+1}=(-1)^3\ {^7\text{C}}_3(2\text{x}^2)^{7-3}\big(\frac{1}{\text{x}}\big)^3$

$=-{^7\text{C}}_3\frac{2^4\text{x}^8}{\text{x}^3}$

$=-560\text{x}^5$

$\text{T}_5=\text{T}_{4+1}=(-1)^4\ {^7\text{C}}_4(2\text{x}^2)^{7-4}\big(\frac{1}{\text{x}}\big)^4$

$=-{^7\text{C}}_4\frac{2^3\text{x}^6}{\text{x}^4}$

$={^7\text{C}}_4\frac{7\times6\times5\times8}{3\times2}\text{x}^2$

$=280\text{x}^2$

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

$\text{N}=8,\ \text{r}=7,\ \text{x}^\frac{3}{2}\text{y}^\frac{1}{2},\ \text{y}=\text{x}^\frac{1}{2}\text{y}^\frac{3}{2},\text{n}=10$

$\text{T}_8=\text{T}_{7+1}=(-1)^7\ {^{10}\text{C}}_7\big(\text{x}^\frac{3}{2}\text{y}^\frac{1}{2}\big)^3\big(\text{x}^\frac{1}{2}\text{y}^\frac{3}{2}\big)^7\\={^{-10}\text{C}}_7\text{x}^\frac{9}{2}\times\text{x}^\frac{7}{2}\times\text{y}^\frac{3}{2}\text{y}^\frac{21}{2}=-120\text{x}^8\text{y}^{12}$

$\text{x}^{-15}\ \text{in}\Big(3\text{x}^2-\frac{\text{a}}{3\text{x}^3}\Big)^{10}$

$(-1)^\text{r}\ {^{40}\text{C}}_\text{r}\text{x}^{40-\text{r}}\Big(\frac{1}{\text{x}^2}\Big)^\text{r}$

$(-1)^\text{r}{^{10}\text{C}}_\text{r}\frac{3^{10-\text{r}}}{3^\text{r}}\text{x}^{20-2\text{r}-3\text{r}}$

$\Rightarrow\text{x}^{20-5\text{r}}=\text{x}^{-15}$

$20-5\text{r}=-15$

$35=5\text{r}$

$\text{r}=7$

$(-1)^7\ {^{10}\text{C}}_7\frac{3^3\text{a}^7}{3^7}$

$-\frac{40}{27}\text{a}^7$

$(1+2\sqrt{\text{x}})^5+(1-2\sqrt{\text{x}})^5$

$=2\big[{^5\text{C}}_0+{^5\text{C}}_2\big(2\sqrt{\text{x}}\big)^2+{^5\text{C}}_4\big(2\sqrt{\text{x}}\big)^4\big]$

$=2\big[1+10\times4\times\text{x}+16\times\text{x}^2\times5\big]$

$=2+80\text{x}+160\text{x}^2$

Fifth term from the end is

(11 - 5 + 1) = 7th term from beginning

$\text{T}_7=\text{T}_{6+1}=(-1)^\text{r}\ {^\text{n}\text{C}}_2\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$

$=(-1)^6\ {^{10}\text{C}}_6(3\text{x})^4\Big(\frac{1}{\text{x}^2}\Big)^6={^{10}\text{C}}_6\times3^4\times\frac{\text{x}^4}{\text{x}^{12}}=\frac{210\times81}{\text{x}^8}=\frac{17010}{\text{x}^8}$

$(\text{a}-2\text{b})^{12}={^{12}\text{C}}_0\text{a}^{12}-{^{12}\text{C}}_1\text{a}^{11}(2\text{b})^1+{^{12}\text{C}}_2\text{a}^{10}(2\text{b})^2\\-{^{12}\text{C}}_3\text{a}^9(2\text{b})^3+\ ......\ +{^{12}\text{C}}_3\text{a}^5(2\text{b})^7+.....$

$=-\frac{12!}{7!5!}\times128$

$=-\frac{12\times11\times10\times9\times8}{5\times4\times3\times2}\times128$

$=-101376$

Here, n = 20 which is an even number so, $\Big(\frac{20}{2}+1\Big)^{\text{th}}$ i.e., 11th term is the middle term.

We know that,

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

$\text{n}=20,\text{r}=10,\text{x}=\frac{2}{3}\text{x},\text{y}=\frac{2}{3\text{x}}$

$\text{T}_{11}=\text{T}_{10+1}=(-1)^{10}\ {^{20}\text{C}}_{10}\big(\frac{2}{3}\text{x}\big)^{10}\big(\frac{3}{2\text{x}}\big)^{10}$

$={^{20}\text{C}}_{10}\frac{2^{10}}{3^{10}}\times\frac{3^{10}}{2^{10}}\times\frac{\text{x}^{10}}{\text{x}^{10}}$

$={^{20}\text{C}}_{10}$

$\text{T}_\text{N}=\text{T}_{\text{r}+1}=(-1)^\text{r}\ {^\text{n}\text{C}}_2\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$

4th term from the end = 7th term from beginning

$\text{N}=7,\text{r}+6,\text{n}=9,\text{x}=\frac{4\text{x}}{5},\text{y}=\frac{5}{2\text{x}}$

$\text{T}_7=\text{T}_{6+1}=(-1)^{6}\ {^9\text{C}}_6\big(\frac{4\text{x}}{5}\big)^3\big(\frac{5}{2\text{x}}\big)^6=\frac{9\times8\times7}{3\times2}\times\frac{4^3\times5^6}{5^3\times2^6}\times\frac{\text{x}^3}{\text{x}^6}\\\frac{9\times8\times7\times5^3}{6\times\text{x}^3}=\frac{9\times8\times7\times125}{6\times\text{x}^3}=\frac{10500}{\text{x}^3}$

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

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

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

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

$\Rightarrow\text{x}^{18-3\text{r}}=\text{x}^9$

$18-3\text{r}=9$

$3\text{r}=9$

$\text{r}=3$

$=(-1)^3\ {^9\text{C}}_3\frac{1}{3^3}$

$=-\frac{9\times8\times7}{3\times2\times9\times3}$

$=\frac{-28}{9}$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of $\bigg(\sqrt\frac{\text{x}}{\text{a}}-\sqrt\frac{\text{a}}{\text{x}}\bigg)^6$ has 7 term Using binomial theorem to expand, we get

$\Big(\sqrt\frac{\text{x}}{\text{a}}-\sqrt\frac{\text{a}}{\text{x}}\Big)^6={^6\text{C}}_0​​\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)^6\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^0\\-{^6\text{C}}_1\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)^5\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^1+{^6\text{C}}_2\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)^4\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^2-{^6\text{C}}_3\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)^3\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^3\\+{^6\text{C}}_4\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)^2\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^4-{^6\text{C}}_5\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^5+{^6\text{C}}_6\Big(\sqrt\frac{\text{x}}{\text{a}}\Big)^0\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^6$

$=\Big(\frac{\text{x}}{\text{a}}\Big)^{\frac{1}{2}\times5}-6\Big(\frac{\text{x}}{\text{a}}\Big)^{\frac{1}{2}\times5}\Big(\frac{\text{a}}{\text{x}}\Big)^{\frac{1}{2}}+15\Big(\frac{\text{x}}{\text{a}}\Big)^{\frac{1}{2}\times4}\Big(\sqrt\frac{\text{a}}{\text{x}}\Big)^{2\times{\frac{1}{2}}}\\-20\Big(\frac{\text{x}}{\text{a}}\Big)^{3\times{\frac{1}{2}}}\Big(\frac{\text{a}}{\text{x}}\Big)^{3\times\frac{1}{2}}+15\Big(\frac{\text{x}}{\text{a}}\Big)^{2\times\frac{1}{2}}\Big(\frac{\text{a}}{\text{x}}\Big)^{4\times\frac{1}{2}}-6\Big(\frac{\text{x}}{\text{a}}\Big)^\frac{1}{2}\Big(\frac{\text{a}}{\text{x}}\Big)^{5\times\frac{1}{2}}+\Big(\frac{\text{a}}{\text{x}}\Big)^{6\times\frac{1}{2}}$

$=\frac{\text{x}^3}{\text{a}^3}-6\frac{\frac{5\ 1}{\text{x}^2\ 2}}{\frac{5}{\text{a}^2}-\frac{1}{2}}+15\times\frac{\text{x}^2\times\text{a}}{\text{a}^2\times\text{x}}-20\times\frac{\frac{3\ 3}{\text{x}^2\ 2}}{\frac{3}{\text{a}^2}-\frac{3}{2}}\\+15\times\frac{\text{x}}{\text{a}}\times\frac{\text{a}^2}{\text{x}^2}-6\times\frac{\frac{1\ 5}{\text{x}^2\ 2}}{\frac{1}{\text{a}^2}-\frac{ 5}{2}}+\frac{\text{a}^3}{\text{x}^3}$

$=\frac{\text{x}^3}{\text{a}^3}-\frac{6\text{x}^2}{\text{a}^2}+\frac{15\text{x}}{\text{a}}-20+\frac{15\text{a}}{\text{x}}-\frac{6\text{a}^2}{\text{x}^2}+\frac{\text{a}^3}{\text{x}^3}$

Here, n = 12 which is an even number

so, $\Big(\frac{12}{2}+1\Big){\text{th}}$ 7th term is the middle term.

Hence, the middle term T₇ = T₆₊₁

^{$\therefore\text{T}_7=\text{T}_{6+1}={^{12}\text{C}}_6\times\big(\frac{\text{a}}{\text{x}}\big)^{12-6}\times(\text{bx})^6$}

$={^{12}\text{C}}_6\big(\frac{\text{a}}{\text{x}}\big)^6\times(\text{bx})^6$

$=\frac{12!}{(12-6)!6!}\times\frac{\text{a}^6}{\text{x}^6}\times\text{b}^6\text{x}^6$

$=\frac{12\times11\times10\times9\times8\times7\times6!}{(6\times5\times4\times3\times2\times1)}\times\text{a}^6\text{b}^6$

$=924\times\text{a}^6\text{b}^6$

$\therefore$ The middle term $=924\times\text{a}^6\text{b}^6.$

7th term from the end = 3^rd term from beginning

$\text{T}_\text{N}=\text{T}_{\text{r}+1}=(-1)^\text{r}\ {^\text{n}\text{C}}_2\text{x}^{\text{n}-\text{r}}\text{y}^\text{r}$

$\text{N}=3,\text{r}=2,\text{n}=8,\text{x}=2\text{x}^2,\text{y}=\frac{3}{2\text{x}}$

$\text{T}_3=\text{T}_{2+1}=(-1)^2\ {^8\text{C}}_2(2\text{x})^6\big(\frac{3}{2\text{x}}\big)^2\\=\frac{8\times7}{2}\times\frac{2^6\times3^2\times\text{x}^{12}}{2^2\times\text{x}^2}=8\times7\times9\times8\times\text{x}^{10}={4032\text{x}^{10}}$

The 6^th, 7^th and 8^th terms in the expansion of $(\text{x}+\text{a})^{\text{n}}$ are  ${^\text{n}}\text{C}_{\text{5}}\text{x}^{\text{n}-5}\text{a}^{5}, \ {^\text{n}}\text{C}_{\text{6}}\text{x}^{\text{n}-6}\text{a}^{6}$ and,

According to the quation, 

 ${^\text{n}}\text{C}_{\text{5}}\text{x}^{\text{n}-5}\text{a}^{5}=112$

${^\text{n}}\text{C}_{\text{6}}\text{x}^{\text{n}-6}\text{a}^{6}=7$

${^\text{n}}\text{C}_{\text{7}}\text{x}^{\text{n}-7}\text{a}^{7}=\frac{1}{7}$

Now,

$\frac{{^\text{n}}\text{C}_{\text{6}}\text{x}^{\text{n}-6}\text{a}^{6}}{{^\text{n}}\text{C}_{\text{5}}\text{x}^{\text{n}-5}\text{a}^{5}}=\frac{7}{112}$

$\Rightarrow \frac{\text{n}-6+1}{6}\text{x}^{-1}\text{a}=\frac{1}{16}$

$\Rightarrow \frac{\text{a}}{\text{x}}=\frac{3}{8\text{n}...40}\ ...(\text{i})$

Also,

$\frac{{^\text{n}}\text{C}_{\text{7}}\text{x}^{\text{n}-7}\text{a}^{7}}{{^\text{n}}\text{C}_{\text{6}}\text{x}^{\text{n}-6}\text{a}^{6}}=\frac{\frac{1}{4}}{7}$

$\Rightarrow \frac{\text{n}-7+1}{7}\text{x}^{-1}\text{a}=\frac{1}{26}$

$\Rightarrow \frac{\text{a}}{\text{x}}=\frac{3}{4\text{n}...24}\ ...(\text{ii})$

From (i) and (ii), we get

$\frac{3}{8\text{n}...40}=\frac{1}{4\text{n}...24}$

$\Rightarrow \frac{3}{2\text{n}...10}=\frac{1}{\text{n}...6}$

$\Rightarrow \text{n}=8$

Putting in equation (i), we get

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

Now,

${^\text{8}}\text{C}_{\text{5}}\text{x}^{\text{8}-5}\big(\frac{\text{x}}{5}\big)^{5}=112$

$\Rightarrow \frac{56\text{x}^{8}}{8^{5}}=112$

$\Rightarrow \text{x}^{8}=4^{8}$

$\Rightarrow \text{x}=4$

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

$\text{a}=\frac{1}{2}$

$\text{a}=3$ and $\text{x}=2$

Let y =1 - 2x, then

$(1-2\text{x}+3\text{x}^2)^3=(\text{y}+3\text{x}^2)^3$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of (y + 3x²)³ has 4 terms. Using binomial theorem to expand, we get

$​​(\text{y}+3\text{x}^2)^3={^3\text{C}}_0\text{y}^3(3\text{x}^2)^0+{^3\text{C}}_1\text{y}^2(3\text{x}^2)^1+{^3\text{C}}_2\text{y}(3\text{x}^2)^2+{^3\text{C}}\text{y}^0(3\text{x}^2)^3$

$=\text{y}^3+3\text{y}^2(3\text{x}^2)+3\text{y}(9\text{x}^2)+(27\text{x}^6)$

Substituting y = 1 - 2x, we get,

$(1-2\text{x}+3\text{x}^2)^3=(1-2\text{x})^3+3(1+4\text{x}^2-4\text{x})(3\text{x}^2)+3(1-2\text{x})(9\text{x}^2)+(27\text{x})^6$

$=1-8\text{x}^3-6\text{x}+12\text{x}^2+9\text{x}^2+36\text{x}^4-36\text{x}^3+27\text{x}^2-54\text{x}^3+27\text{x}^6$

$=1-6\text{x}+21\text{x}^2-44\text{x}^3+63\text{x}^4-54\text{x}^5+27\text{x}^6$

$\Big(3-\frac{\text{x}^{3}}{6}\Big)^{7}$

Hence n = 7, which is even therefore it has 11 terms.

Middle term is $\Big(\frac{7+1}{2}\Big)$ and $\Big(\frac{7+1}{2}+1\Big)=4^{*},5^{*}$

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

$\text{T}_{\text{4}}=\text{T}_{\text{3}+1}=(-1)^{\text{3}}\ {^\text{7}}\text{C}_{\text{3}}(3)^{7-3}\Big(\frac{\text{x}^{3}}{6}\Big)^{3}$

$=-\frac{7!}{3!4!}\times3^{4}\times\frac{\text{x}^{3}}{6^{3}}$

$=-\frac{7\times6\times5}{3\times2\times1}\times81\times\frac{\text{x}^{9}}{216}$

$=-\frac{105}{8}\text{x}^{9}$

And,

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

$\text{T}_{\text{5}}=\text{T}_{\text{4}+1}=(-1)^{\text{4}}\ {^\text{7}}\text{C}_{\text{4}}(3)^{7-4}\Big(\frac{\text{x}^{3}}{6}\Big)^{4}$

$=-\frac{7!}{4!3!}\times3^{3}\times\frac{\text{x}^{12}}{6^{4}}$

$=-\frac{7\times6\times5}{3\times2\times1}\times27\times\frac{\text{x}^{12}}{1296}$

$=-\frac{35}{48}\text{x}^{12}$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of $\Big(\text{ax}-\frac{\text{b}}{\text{x}}\Big)^6$has 7 term Using binomial theorem to expand, we get

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

$=\text{a}^6\text{x}^6-6\text{a}^5\text{x}^5\frac{\text{b}}{\text{x}}+15\text{a}^4\text{x}^4\frac{\text{b}^2}{\text{x}^2}\\-20\text{a}^3\text{b}^3+15\text{a}^2\frac{\text{b}^4}{\text{x}^2}-6\text{a}\frac{\text{b}^5}{\text{x}}^4+\frac{\text{b}^6}{\text{x}^6}$

$=\text{a}^6\text{x}^6-6\text{a}^5\text{x}^4\text{b}\frac{\text{b}}{\text{x}}+15\text{a}^4\text{b}^2\text{x}^2-20\text{a}^3\text{b}^3+15\frac{\text{a}^2\text{b}^4}{\text{x}^2}-6\frac{\text{ab}^5}{\text{x}^4}+\frac{\text{b}^6}{\text{x}^6}$

$\Big({\text{x}}-\frac{1}{\text{x}}\Big)^{10}$

Here, n = 10, which is even, $\therefore$ it has 11 terms

$\therefore$ middle term is $\Big(\frac{\text{n}}{2}+1\Big)=6^\text{th}\ \text{term}$

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

$\text{T}_6=\text{T}_{5+1}=(-1)^5\ {^{10}\text{C}}_5(\text{x})^{10-5}\Big(\frac{1}{\text{x}}\Big)^5$

$=\frac{-10​​\times9\times89\times7\times6}{5\times4\times3\times2}\times\frac{\text{x}^5}{\text{x}^5}$

$=-3\times2\times\times6$

$=-252$

$(1+\text{x})^{43}$

$\left(\begin{array}{c}43\\ 2\text{r}\end{array}\right)=\left(\begin{array}{c}43\\ \text{r+1}\end{array}\right)$

$2\text{r}+\text{r}+1=43$

$3\text{r}=42$

$\text{r}=14$

Let y = x + 1, then

$\Big(\text{x}+1-\frac{1}{\text{x}}\Big)^3=\Big(\text{y}-\frac{1}{\text{x}}\Big)^3$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of $\Big(\text{y}-\frac{1}{\text{x}}\Big)^3$ has 4 terms. Using binomial theorem to expand, we get

$\Big(\text{y}-\frac{1}{\text{x}}\Big)^3={^3\text{C}}_0\text{y}^3\Big( \frac{1}{\text{x}}\Big)^0-{^3\text{C}}_1\text{y}^2\Big(\frac{1}{\text{x}}\Big)+{^3\text{C}}_2\text{y}\Big(\frac{1}{\text{x}}\Big)^2-{^3\text{C}}_3\text{y}^0\Big(\frac{1}{\text{x}}\Big)^3$

$=\text{y}^3-3\text{y}^2\times\frac{1}{\text{x}}+3\text{y}\times\frac{1}{\text{x}^2}-\frac{1}{\text{x}^3}$

Putting y = x + 1, we get

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

$=\text{x}^3+1+3\text{x}^2+3\text{x}-3\text{x}-\frac{3}{\text{x}}-6+\frac{3}{\text{x}}+\frac{3}{\text{x}^2}-\frac{1}{\text{x}^3}$

$=\text{x}^3+3\text{x}^2-5+\frac{3}{\text{x}^2}-\frac{1}{\text{x}^3}$

The expansion of (x + y)ⁿ has n + 1 terms so the expansion of (1 - 3x)⁷ has 8 term Using binomial theorem to expand, we get

$(1-3\text{x})^7={^7\text{C}}_0(1)^7(3\text{x})^0-{^7\text{C}}_1(3\text{x})+{^7\text{C}}_2(3\text{x})^2-{^7\text{C}}_3(3\text{x})^3\\+{^7\text{C}}_4(3\text{x})^4-{^7\text{C}}_5(3\text{x})^5-{^7\text{C}}_6(3\text{x})^6$

$=1-21\text{x}+21\times9\text{x}^2-35\times3^3\text{x}^3+35\times3^4\text{x}^4-\\21\times3^5\text{x}^5+7\times3^6\text{x}^6-3^7\text{x}^7$

$=1-21\text{x}+189\text{x}^2-945\text{x}^3+2835\text{x}^4-5103\text{x}^5+5103\text{x}^6-2187\text{x}^7$

The expansion of (x + y)ⁿ has n + 1 term so, the expansion of (2x - 3y)⁴ has 5 terms.

Using binomial theorem, we have

$(2\text{x}-3\text{y})^4={^4\text{C}_0}(2\text{x})^4(3\text{y})^0-{^4\text{C}}_1(2\text{x})^3(3\text{y})^1\\+{^4\text{C}}_2(2\text{x})^2(3\text{y})^2-{^4\text{C}}_3(2\text{x})^1(3\text{y})^3+{^4\text{C}}_4(2\text{x})^0(3\text{y})^4$

$=2^4\text{x}^4-4\times2^3\times3\text{x}^3\text{y}+6\times2^2\times3^2\times\text{x}^2\text{y}^2\\-4\times2\times3^3\times\text{xy}^3+\text{3}^4\text{y}^4$

$=16\text{x}^4-96\text{x}^3\text{y}+216\text{x}^2\text{y}^2-216\text{xy}^3+81\text{y}^4$

The expansion of (x + y)ⁿ has n + 1 term so, the expansion of (2x + 3y)⁵ has 6 terms.

Using binomial theorem, we have

$(2\text{x}+3\text{y})^5={^5\text{C}}(2\text{x})^5(3\text{y})^0+{^5\text{C}}_1(2\text{x})^4(3\text{y})^1+{^5\text{C}}_2(2\text{x})^3(3\text{y})^2\\+{^5\text{C}}_3(2\text{x})^2(3\text{y})^3+{^5\text{C}}_4(2\text{x})(3\text{y})^4+{^5\text{C}}_5(2\text{x})^0(3\text{y})^5$

$=2^5\text{x}^5+5\times2^4\times3\times\text{x}^4\times\text{y}+10\times2^3\times3^2\times\text{x}^3\times\text{y}^2+10\times2^2\\\times3^3\times\text{x}^2\times\text{y}^3+5\times2\times3^4\times\text{x}\times\text{y}+3^5\text{y}^5$

$=32\text{x}^5+240\text{x}^4\text{y}+720\text{x}^3\text{y}^2+1080\text{x}^2\text{y}^3+810\text{xy}^4+243\text{y}^5$

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

Then,

$\text{T}_{3}=\text{a}=​​​​​{^\text{n}}\text{C}_{\text{2}} \text{x}^{\text{n}-2} \alpha^{2}=\frac{\text{n}(\text{n}-1)}{2}\text{x}^{\text{n}-2}\alpha^{2}$

$\text{T}_{4}=\text{b}=​​​​​{^\text{n}}\text{C}_{\text{3}} \text{x}^{\text{n}-3} \alpha^{3}=\frac{\text{n}(\text{n}-1)(\text{n}-2)}{6}\text{x}^{\text{n}-3}\alpha^{3}$

$\text{T}_{5}=\text{c}=​​​​​{^\text{n}}\text{C}_{\text{4}} \text{x}^{\text{n}-4} \alpha^{4}=\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)}{24}\text{x}^{\text{n}-4}\alpha^{4}$

$\text{T}_{6}=\text{d}=​​​​​{^\text{n}}\text{C}_{\text{5}} \text{x}^{\text{n}-5} \alpha^{5}=\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)(\text{n}-4)}{120}\text{x}^{\text{n}-5}\alpha^{5}$

Now,

$\frac{\text{T}_{4}}{\text{T}_{3}}=\frac{\text{b}}{\text{a}}=\frac{\frac{\text{n}(\text{n}-1)(\text{n}-2)\text{x}^{\text{n}-3}\alpha^{3}}{6}}{\frac{\text{n}(\text{n}-1)}{2}\text{x}^{\text{n}-2}\alpha^{2}}=\Big(\frac{\text{n}-2}{3}\Big).\frac{\alpha}{\text{x}}\ ...(\text{i})$

Similarly,

$\frac{\text{T}_{5}}{\text{T}_{4}}=\frac{\text{c}}{\text{b}}=\frac{\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)\text{x}^{\text{n}-4}\alpha^{4}}{24}}{\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)}{6}\text{x}^{\text{n}-3}\alpha^{3}}=\Big(\frac{\text{n}-3}{4}\Big).\frac{\alpha}{\text{x}}\ ...(\text{ii})$

and,

$\frac{\text{T}_{6}}{\text{T}_{5}}=\frac{\text{d}}{\text{c}}=\frac{\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)(\text{n}-4)\text{x}^{\text{n}-5}\alpha^{5}}{120}}{\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)}{24}\text{x}^{\text{n}-4}\alpha^{4}}=\Big(\frac{\text{n}-4}{5}\Big).\frac{\alpha}{\text{x}}\ ...(\text{iii})$

Again, dividing (i) by (ii) and (ii) by (iii), we get

$\frac{\frac{\text{b}}{\text{a}}}{\frac{\text{c}}{\text{b}}}=\frac{\Big(\frac{\text{n}-2}{3}\big).\frac{\alpha}{\text{x}}}{\Big(\frac{\text{n}-3}{4}\Big).\frac{\alpha}{\text{x}}}=\frac{4(\text{n}-2)}{3(\text{n}-3)}$

$\Rightarrow \frac{\text{b}^{2}}{\alpha\text{c}}=\frac{4(\text{n}-2)}{3(\text{n}-3)}\ ...(\text{iv})$

and,

$\frac{\frac{\text{c}}{\text{b}}}{\frac{\text{d}}{\text{c}}}=\frac{\Big(\frac{\text{n}-3}{4}\big).\frac{\alpha}{\text{x}}}{\Big(\frac{\text{n}-4}{5}\Big).\frac{\alpha}{\text{x}}}=\frac{5(\text{n}-3)}{4(\text{n}-4)}$

$\Rightarrow \frac{\text{c}^{2}}{\text{bd}}=\frac{5(\text{n}-3)}{4(\text{n}-4)}\ ...(\text{v})$

Now subtracting 1 from both sides of quation (iv) and (v) as:

$\Rightarrow \frac{\text{b}^{2}}{\alpha\text{c}}-1=\frac{4(\text{n}-2)}{3(\text{n}-3)}-1$

$\Rightarrow \frac{\text{b}^{2}-\text{ac}}{\alpha\text{c}}=\frac{\text{n}+1}{3(\text{n}-3)}\ ...(\text{vi})$

$\Rightarrow \frac{\text{c}^{2}}{\text{bd}}-1=\frac{5(\text{n}-3)}{4(\text{n}-4)}-1$

$\Rightarrow \frac{\text{c}^{2}-\text{bd}}{\text{bd}}=\frac{\text{n}+1}{4(\text{n}-4)}\ ...(\text{vii})$

Again, on diving (vi) by (vii), we get

$\frac{\frac{\text{b}^{2}-\text{ac}}{\text{ac}}}{\frac{\text{c}^{2}-\text{bd}}{\text{bd}}}=\frac{\frac{\text{n}+1}{3(\text{n}-3)}}{\frac{\text{n}+1}{4(\text{n})-4}}$

$\Rightarrow \frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}\times\frac{\text{bd}}{\text{ac}}=\frac{4(\text{n}-4)}{3(\text{n}-3)}\ ...(\text{viii})$

On multipiying (v) by (viii), we get

$\Rightarrow \frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}\times\frac{\text{bd}}{\text{ac}}\times\frac{\text{c}^{2}}{\text{bd}}=\frac{4(\text{n}-4)}{3(\text{n}-3)}\times\frac{5(\text{n}-3)}{4(\text{n}-4)}$

$\Rightarrow \frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}\times\frac{\text{c}}{\text{a}}=\frac{5}{3}$

$\Rightarrow \frac{\text{b}^{2}-\text{ac}}{\text{c}^{2}-\text{bd}}=\frac{5\text{a}}{3\text{c}}\ \dots(\text{IX})$

By Equation (IX), it is proved that,

L.H.S = R.H.S

Any term in the expansion of $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{12}\ \text{is}$

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

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

$={^{12}\text{C}}_\text{r}\text{x}^{24-2\text{r}}\text{x}^{-12}$

$\text{x}^{12-2\text{r}}=\text{x}^{-1}$

$12-2\text{r}=-1$

$2\text{r}=13$

$\text{r}=\frac{13}{2}$

r can not be a fraction, therefore there is no term in the expansion of $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{12}$ having the term x^-1.

$\text{T}_{\text{r}+1}={^\text{10}}\text{C}_{\text{r}}\Big(\sqrt{\frac{\text{x}}{3}}\Big)^{10-\text{r}}\Big(\frac{3}{2\text{x}^{2}}\Big)^{\text{r}}$

$={^\text{10}}\text{C}_{\text{r}}\ \text{x}^{5-\frac{\text{r}}{2}2\text{r}}\ 3^\text{r}\times3^{5+\frac{\text{r}}{2}}\times2^{-\text{r}}$

Independent of x = x⁰

$\text{x}\frac{10-\text{r}-4\text{r}}{\text{r}}=\text{x}^{0}$

$10-5\text{r}=0$

$\text{r}=2$

$\text{T}_{3}={^\text{10}}\text{C}_{\text{2}}\ 3^\text{-2}\ 2^{-2}$

$={^\text{10}}\text{C}_{\text{2}}\ 3^\text{-2}\ 2^{-2}$

$=\frac{10!}{2!8!}\times\frac{1}{63}$

$=\frac{10\times9}{2\times36}=\frac{5}{4}$

$\text{T}_{\text{r+1}}=(-1)^{\text{r}} \ {^\text{15}}\text{C}_{\text{r}}\big(3\text{x}^{2})^{15-\text{r}}\Big(\frac{2}{\text{x}^{2}}\Big)^{\text{r}}$

$=(-1)^{\text{r}} \ {^\text{15}}\text{C}_{\text{r}}\big(3)^{15-\text{r}}\ 2^{\text{r}}\ \text{x}^{15-\text{r}-2\text{r}}$

Term independent of x = x⁰

$\Rightarrow \text{x}^{51-3\text{r}}=\text{x}^{0}$

$\Rightarrow 15-3\text{r}=0$

$\Rightarrow \text{r}=5$

$\text{T}_{11}=(-1)^{\text{5}} \ {^\text{15}}\text{C}_{\text{5}}\big(3)^{10}\times 2^{\text{5}}$

$=-\frac{15\times14\times13\times12\times11}{120}\text{3}^{\text{10}}2^{5}$

$=-3003\times\big(3)^{10}\times 2^{\text{5}}$

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

Here, n is odd, $\therefore(1-2\text{x}+\text{x}^2)^\text{n}$ has n + 1 = even term

$\therefore$ middle term is $\Big(\frac{\text{n}+1}{2}\Big)^\text{th}$ term

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

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

$=\frac{\text{n}!}{\frac{\text{n}}{2}!\frac{\text{n}}{2}!}(1-2\text{x})^{\frac{\text{n}}{2}}\text{x}^{\frac{2\text{n}}{2}}$

$=\frac{(2\text{n})!}{(\text{n}!)}(-1)^\text{n}\text{x}^\text{n}\big[\because(1-\text{x})^\text{n}=1-\text{nx}\big]$

$(\sqrt[3]{\text{x}}-\sqrt[3]{\text{a}})^6$

$=\Big(\frac{6}{0}\Big)\big(\sqrt[3]{\text{x}}\big)^6\big(-\sqrt[3]{\text{a}}\big)^0+\Big(\frac{6}{1}\Big)\big(\sqrt[3]{\text{x}}\big)^5\big({-}\sqrt[3]{\text{a}}\big)^1+\Big(\frac{6}{2}\Big)\big(-\sqrt[3]{\text{x}}\big)^4\big(\sqrt[3]{\text{a}}\big)^2\\\Big(\frac{6}{3}\Big)\big(\sqrt[3]{\text{x}}\big)^3\big(-\sqrt[3]{\text{a}}\big)^0+\Big(\frac{6}{4}\Big)\big(\sqrt[3]{\text{x}}\big)^2\big(-\sqrt[3]{\text{a}}\big)^4+\Big(\frac{6}{5}\Big)\big(\sqrt[3]{\text{x}}\big)^1\big(-\sqrt[3]{\text{a}}\big)^5\\\Big(\frac{6}{6}\Big)\big(\sqrt[3]{\text{x}}\big)^0\big(-\sqrt[3]{\text{a}}\big)^6$

$=\text{x}^2-6\text{x}^\frac{5}{3}\text{a}^\frac{2}{3}+15\text{x}^\frac{4}{3}\text{a}^\frac{2}{3}-20\text{ax}+15\text{x}^\frac{2}{3}\text{a}^\frac{4}{3}-6\text{x}^\frac{1}{3}\text{a}^\frac{5}{3}+\text{a}^2$