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

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(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}^{\

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$2n^2 + 2n + 1$

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(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

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