MATHS 1 Marks Question

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

Here k = 2n, which is number.

So, $\Big(\frac{2\text{n}}{2}+1\Big)\text{th}$ term i.e. (n + 1)th term is the middle term.

Coefficient of (r + 1)th term in the binomial expansion of $(1+\text{x})^{\text{k}}$ is ${^\text{k}}\text{C}_{\text{r}}.$ 

Coefficient of (n + 1)th term in the binomial expansion of $(1+\text{x})^{\text{2}\text{n}}$ is ${^\text{2n}}\text{C}_{\text{n}}.$

The total number of terms are 101 of which 50 terms get cancelled.

Hence, the total number of terms in the expansion of $(\text{x}+\text{a})^{100}+(\text{x}-\text{a})^{100}$ is 51.

Coefficients of x^p in the expansion of $(1+\text{x})^{\text{p}+\text{q}}$ is ${^\text{q}}\text{C}_{\text{p}}.$

Coefficients of x^q in the expansion of $(1+\text{x})^{\text{p}+\text{q}}$ is ${^\text{q}}\text{C}_{\text{p}}.$

Now,

$\frac{{^\text{p+q}}\text{C}_{\text{p}}}{{^\text{p+q}}\text{C}_{\text{q}}}=\frac{\frac{(\text{p+q})!}{\text{p!}\text{q}!}}{{\frac{(\text{p+q})!}{\text{q!}\text{p}!}}}=1$

Hence, the ratio of the coefficients of x^p and x^q in the expansion of $(1+\text{x})^{\text{p}+\text{q}}$ is 1 : 1.

$3^{400}=(9)^{200}$

$=(10-1)^{200}$

$={^\text{200}}\text{C}_{\text{0}}(10)^{200}+{^\text{200}}\text{C}_{\text{1}}(10)^{199}(-1)^{1}.....\\+{^\text{200}}\text{C}_{\text{108}}(10)^{2}(-1)^{198}+{^\text{200}}\text{C}_{\text{199}}(10)^{1}(-1)^{199}+{^\text{200}}\text{C}_{\text{200}}(-1)^{200}$

$=100\Big[(10)^{198}+{^\text{200}}\text{C}_{\text{1}}(10)^{197}(-1)^{1}+......+{^\text{200}}\text{C}_{\text{198}}(-1)^{198}\Big]+200(10)^{199}+(-1)^{200}$

$=100\Big[(10)^{198}-{^\text{200}}\text{C}_{\text{1}}(10)^{197}+......+{^\text{200}}\text{C}_{\text{198}}-2(10)\Big]+1$

$=100(\text{a natural number})+1$

Hence, last two digits of the number 3⁴⁰⁰ is 01.

Putting x = 1 and -1 in $(1-\text{x}+\text{x}^{2})^{\text{n}}=\text{a}^{0}+\text{a}_{1}\text{x}+\text{a}_{2}\text{x}^{2}+...+\text{a}_{2\text{n}}\text{x}^{2\text{n}}$

we get,

$1=\text{a}_{0}+\text{a}_{1}+\text{a}_{2}+....+\text{a}_{2\text{n}}\ ...(\text{1})$ and,

$3^{\text{n}}=\text{a}_{0}-\text{a}_{1}+\text{a}_{2}-....+\text{a}_{2\text{n}}\ ...(\text{2})$

Adding (i) and (ii), we get

$3^{\text{n}}+1=2(\text{a}_{0}+\text{a}_{2}+...+\text{a}_{2\text{n}})$

Hence, the value of $\text{a}_{0}+\text{a}_{2}+\text{a}_{4}+..\text{a}_{2\text{n}}$ is $\frac{3^{\text{n}}+1}{2}.$

$(1-3\text{x}+\text{x}^{2})^{111}$

$={^\text{n}}\text{C}_{\text{0}}(1)^{111}+{^\text{111}}\text{C}_{\text{1}}(1)^{100}(-3\text{x}+\text{x}^{2})+{^\text{111}}\text{C}_{\text{2}}(1)^{109}\$-3\text{x}+\text{x}^{2})^{2+....... +{^\text{111}}\text{C}_{\text{111}}}(-3\text{x}+\text{x}^{2})^{111}$

Let x = 1 on both sides,

$(1-3+1)^{111}$

$={^\text{n}}\text{C}_{\text{0}}(1)^{111}+{^\text{111}}\text{C}_{\text{1}}(1)^{100}(-3+1)+{^\text{111}}\text{C}_{\text{2}}(1)^{109}\$-3+1)^{2+....... +{^\text{111}}\text{C}_{\text{111}}}(-3+1)^{111}$

The sum of the coefficients is (-1)¹¹¹ = -1.

We have,

$(1-3\text{x}+3\text{x}^{2}-\text{x}^{3})^{8}=\Big[(1+\text{x}^{3})\Big]^{8}=(1-\text{x})^{24}$

So, there are 25 terms in the expansion of $(1-3\text{x}+3\text{x}^{2}-\text{x}^{3})^{8}.$

Number of terms in the expansion $(\text{x}+\text{y})^{\text{n}}+(\text{x}-\text{y})^{\text{n}}$ where n is even $=\big(\frac{\text{n}}{2}+1\big)$ 

Thus, we have

Number of terms in the given expansion $=\Big(\frac{10}{2}+1\Big)=6$

We have,

$(\text{a}+\text{b}+\text{c})^{\text{n}}=\big[\text{a}+(\text{b}+\text{c})\big]^{\text{n}}$

$=\text{a}^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}\text{a}^{\text{n}-1}(\text{b}+\text{c})^{1}+{^\text{n}}\text{C}_{\text{2}}\text{a}^{\text{n}-2}(\text{b}+\text{c})^{2}....+{^\text{n}}\text{C}_{\text{n}}(\text{b}+\text{c})^{\text{n}}$

Further, expanding each term of R.H.S., we note that,

First term consists of 1 term.

Second term on simplification gives 2 terms.

Third term on expansion gives 3 terms.

Similarly, fourth term on expansion gives 4 terms and so on.

$\therefore$ The total number of term $=1+2+3+...+(\text{n}+1)=\frac{(\text{n}+1)(\text{n}+2)}{2}.$

The coefficients of xⁿ in $(1+\text{x})^{2\text{n}}$ is twice the coefficient of xⁿ in $(1+\text{x})^{2\text{n}-1}.$

$\therefore$ a = 2b

$\Big(\frac{2\text{x}^{2}}{3}+\frac{3}{2\text{x}^{2}}\Big)^{10}$

Hence n = 10, which is even number.

So, $\Big(\frac{10}{2}+1\Big)\text{th}$ term i.e., 6th term is the middle term.

$\text{T}_{6}=\text{T}_{5+1}={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{10-5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$

$={^\text{10}}\text{C}_{\text{5}}\Big(\frac{2\text{x}^{2}}{3}\Big)^{5}\Big(\frac{3}{2\text{x}^{2}}\Big)^{5}$

$=252$

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

Here, n is an odd number.

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

 Now, we have

$\text{T}_{\text{n}}=\text{T}_{\text{n}-1+1}$

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

And,

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

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

$\therefore$ The coefficient of two middle terms are ${^\text{2n-1}}\text{C}_{\text{n}-1}$ and ${^\text{2n-1}}\text{C}_{\text{n}}.$

Now,

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

Hence, the sum of the coefficients of two middle terms in the binomial expansion of $(1+\text{x})^{2\text{n}-1}$ is ${^\text{2n}}\text{C}_{\text{n}}.$

$(1-3\text{x}+10\text{x}^{2})^\text{n}$

$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}-1}(-3\text{x}+10\text{x}^{2})+....+{^\text{n}}\text{C}_{\text{n}}(-3\text{x}+10\text{x}^{2})^{\text{n}}$

Let x = 1 on both sides.

$(1-3+10)^{\text{n}}$

$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}}(-3+10)+...+{^\text{n}}\text{C}_{\text{n}}(-3+10)^{\text{n}}$

$\therefore 8^{\text{n}}=\text{a}$

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

$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}-1}(\text{x}^{2})+.....{^\text{n}}\text{C}_{\text{n}}(\text{x}^{2})^{\text{n}}$

Let x = 1 on both sides.

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

$={^\text{n}}\text{C}_{\text{0}}(1)^{\text{n}}+{^\text{n}}\text{C}_{\text{1}}(1)^{\text{n}-1}(\text{x}^{2})+.....{^\text{n}}\text{C}_{\text{n}}(\text{x}^{2})^{\text{n}}$

$\therefore 2^{\text{n}}=\text{b}$

$\therefore \text{a}=\text{b}^{3}$

Coefficients of xⁿ in the expansion of $(1+\text{x})^{2\text{n}}$ is ${^\text{2n}}\text{C}_{\text{n}}=\text{a}.$

Coefficients of xⁿ in the expansion of $(1+\text{x})^{2\text{n}-1}$ is ${^\text{2n-1}}\text{C}_{\text{n}}=\text{b}.$

Now,

$\frac{\text{a}}{\text{b}}=\frac{​​{^\text{2n}}\text{C}_{\text{n}}}{{^\text{2n-1}}\text{C}_{\text{n}}}$

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

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

$=2$

Hence, $\frac{\text{a}}{\text{b}}=2.$

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

Hence n = 10, which is even number.

So, $\Big(\frac{10}{2}+1\big)\text{th}$ term i.e., 6th term is the middle term.

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

$={^\text{10}}\text{C}_{\text{5}}$

In the binomial expansion of $(1+\text{x})^{\text{m}+\text{n}}$ coefficients of x^m and xⁿ are equal.

$\therefore$ a = b

In the binomial expansion of $(\text{a}+\text{b})^{\text{n}},$ total number of term will be (n + 1).

Now, $\Big[(2\text{x}+\text{y}^{3})^{4}\Big]^{7}=(2\text{x}+\text{y})^{28}$

Therefore, in the expansion of $\Big[(2\text{x}+\text{y}^{3})^{4}\Big]^{7},$ total number of term will be 28 + 1 = 29.

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

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

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

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

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

This term is independent of x, if

9 - 3r = 0

⇒ r = 3

So, (3 + 1)th i.e., 4th term is independent of x.