Arrange expansion of left(x^{1 / 2}+frac{1}{2 x^{1 / 4}}right)^n in decreasing powers of x .Suppose

English

Gujarati

Hindi

7.Binomial Theorem

normal

Arrange the expansion of $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$ in decreasing powers of $x$.Suppose the coeff icients of the first three terms form an arithmetic progression. Then, the number of terms in the expansion having integer power of $x$ is

$1$

$2$

$3$

more than $3$

(KVPY-2010)

Solution

(c)

We have, $\left(x^{1 / 2}+\frac{1}{2 x^{1 / 4}}\right)^n$

$T_{r+1}={ }^n C_r\left(x^{1 / 2}\right)^{n-r} \frac{1}{\left(2 x^{\frac{1}{4}}\right)^r}$

$T_{r+1}={ }^n C_r x{ }^{n-r}-{ }^r 4 \cdot 2^{-r}$

$T_{r+1}={ }^n C_r \frac{2 n-3 r}{4} \cdot 2^{-r}=\frac{{ }^n C_r}{2^r} x^{\frac{2 n-3 r}{4}}$

Given $T_1, T_2, T_3$ are in AP.

$\therefore 2 T_2=T_1+T_3$

$\frac{2^n C_1}{2}={ }^n C_0+\frac{{ }^n C_2}{2^2}$

$\Rightarrow { }^n C_1={ }^n C_0+\frac{n(n-1)}{2 \cdot 2^2}$

$\Rightarrow \quad n=1+\frac{n(n-1)}{8}$

$\Rightarrow \quad n-1=\frac{n(n-1)}{8}$

$\Rightarrow n-1=0 \text { or } n=8$

When $n=8$,

$\frac{2 n-3 r}{4}=\frac{16-3 r}{4}$ is integer

If $\quad r=0,4,8$

Standard 11

Mathematics

The coefficient of ${x^{32}}$ in the expansion of ${\left( {{x^4} – \frac{1}{{{x^3}}}} \right)^{15}}$ is

easy

The term independent of $x$ in the expansion of ${(1 + x)^n}{\left( {1 + \frac{1}{x}} \right)^n}$ is

hard

If coefficient of ${(2r + 3)^{th}}$ and ${(r – 1)^{th}}$ terms in the expansion of ${(1 + x)^{15}}$ are equal, then value of r is

medium

The term independent of $x$ in the expansion of ${\left( {\sqrt {\frac{x}{3}} + \frac{3}{{2{x^2}}}} \right)^{10}}$ will be

easy

(IIT-1965)

For some $n \neq 10$, let the coefficients of the $5^{\text {th }}, 6^{\text {th }}$ and $7^{\text {th }}$ terms in the binomial expansion of $(1+x)^{\text {n+4 }}$ be in $A.P.$ Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is :