Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

$\frac{(1+x)}{(1+x^2)(x-1)}=\frac{x}{(1+x^2)}+\frac{1}{(1-x)}$

The general term for ${(1-x)}^{-1}=1+x+x^2+x^3+.........$

${(1-x)}^{-1}=\sum _0^{\infty }{x}^n.........\left(i\right)$

Replacing $x$ by $-x^2$

$\frac{1}{1+x^2}=1-x^2+x^4-x^6.......\frac{1}{1+x^2}=\sum _0^{\infty }{{(-1)}^{n}x}^{2n}$

Multiplying both sides by $x$

$\frac{x}{1+x^2}=\sum _0^{\infty }{{(-1)}^{n}x}^{2n+1}....\left(ii\right)$

From (i) clearly the cofficient of $x^{2012}$ is $1$ and there is no term with even cofficient in $\left(ii\right)$

So the cofficient of $x^{2012}$ in the given expression is $1$