The coefficient of x^{2012} in frac{1+x}{(1+x | vedclass

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(A) Let $f(x) = \frac{1+x}{(1+x^2)(1-x)}$.Using partial fractions,we write $\frac{1+x}{(1+x^2)(1-x)} = \frac{Ax+B}{1+x^2} + \frac{C}{1-x}$.$1+x = (Ax+

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$1$

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(A) Let $f(x) = \frac{1+x}{(1+x^2)(1-x)}$.Using partial fractions,we write $\frac{1+x}{(1+x^2)(1-x)} = \frac{Ax+B}{1+x^2} + \frac{C}{1-x}$.$1+x = (Ax+B)(1-x) + C(1+x^2) = Ax - Ax^2 + B - Bx + C + Cx^2$.Comparing coefficients:$x^2: -A + C = 0 \implies A = C$.$x^1: A - B = 1$.$x^0: B + C = 1$.Substituting $A=C$ into $A-B=1$ gives $C-B=1$. Adding $B+C=1$ gives $2C=2$,so $C=1$ and $A=1$. Then $B=0$.Thus,$f(x) = \frac{x}{1+x^2} + \frac{1}{1-x} = x(1+x^2)^{-1} + (1-x)^{-1}$.Using the binomial expansion $(1+z)^{-1} = \sum_{n=0}^{\infty} (-1)^n z^n$ and $(1-z)^{-1} = \sum_{n=0}^{\infty} z^n$:$f(x) = x \sum_{n=0}^{\infty} (-1)^n x^{2n} + \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} (-1)^n x^{2n+1} + \sum_{n=0}^{\infty} x^n$.We want the coefficient of $x^{2012}$.The first sum only contains odd powers of $x$,so the coefficient of $x^{2012}$ is $0$.The second sum is $\sum_{n=0}^{\infty} x^n$,so the coefficient of $x^{2012}$ is $1$.Total coefficient $= 0 + 1 = 1$.

The coefficient of $x^{2012}$ in $\frac{1+x}{(1+x^2)(1-x)}$ is

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