The coefficient of x^n in frac{(1 + x)^2}{(1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We have $\frac{(1 + x)^2}{(1 - x)^3} = (1 + 2x + x^2)(1 - x)^{-3}$.Using the binomial expansion for negative indices,$(1 - x)^{-3} = \sum_{r=0}^{\

Vedclass · Accepted Answer

$2n^2 + 2n + 1$

Vedclass · Answer

(B) We have $\frac{(1 + x)^2}{(1 - x)^3} = (1 + 2x + x^2)(1 - x)^{-3}$.Using the binomial expansion for negative indices,$(1 - x)^{-3} = \sum_{r=0}^{\infty} \binom{r+3-1}{r} x^r = \sum_{r=0}^{\infty} \binom{r+2}{2} x^r$.Thus,the expression is $(1 + 2x + x^2) \sum_{r=0}^{\infty} \binom{r+2}{2} x^r$.The coefficient of $x^n$ is obtained by taking the sum of coefficients from the product:$= 1 \cdot \binom{n+2}{2} + 2 \cdot \binom{n-1+2}{2} + 1 \cdot \binom{n-2+2}{2}$$= \binom{n+2}{2} + 2 \binom{n+1}{2} + \binom{n}{2}$$= \frac{(n+2)(n+1)}{2} + 2 \frac{(n+1)n}{2} + \frac{n(n-1)}{2}$$= \frac{1}{2} [n^2 + 3n + 2 + 2n^2 + 2n + n^2 - n]$$= \frac{1}{2} [4n^2 + 4n + 2] = 2n^2 + 2n + 1$.

The coefficient of $x^n$ in $\frac{(1 + x)^2}{(1 - x)^3}$ is

Similar Questions

What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y)^{-5}$,when $\left|\frac{y}{x}\right| < 1$ ?

The set of all real values of $x$ for which the expansion of $\left(125 x^2-\frac{27}{x}\right)^{-2/3}$ is valid,is

The coefficient of $x^3$ in the expansion of $(1-\frac{3}{4} x)^{\frac{1}{2}}$ is

The sum of the series $\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}-\ldots$

If $x > \sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x^{-2}$,then the coefficient of $x^{-8}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module