The problem is to find coefficient of $x^n$ using binomial theorem for rational index in the expansion of $$\frac{1}{1-x+x^2-x^3}.$$ In my book the answer is given as $$\frac14+\frac{n+1}{2}+\frac{(-1)^n}{4}.$$ I think the answer in book is wrong.
I will be grateful if anyone check and tell this answer is correct or not with explanation. thanks
Perhaps somewhat easier:
$$1-x+x^2-x^3=(1-x)(1+x^2)\implies\frac1{1-x+x^2-x^3}=\frac1{(1-x)(1+x^2)}$$
Now simple fractions:
$$\frac1{(1-x)(1+x^2)}=\frac A{1-x}+\frac{Bx+C}{1+x^2}\implies1=A(1+x^2)+(Bx+C)(1-x)$$
Comparing coefficients for different values of $\;x\;$ , say for $\;x=1\;$ and coefficients of different powers of $\;x\;$ , we get:
$$\begin{align*}&x=1: \implies&1=2A\implies \color{red}{A=\frac12}\\{}\\&x^2: \implies&0=A-B\implies \color{red}{B=\frac12}\\{}\\&x:\implies&0=B-C\implies \color{red}{C=\frac12}\end{align*}$$
and we get
$$\frac1{(1-x)(1+x^2)}=\frac12\left(\frac1{1-x}+\frac{1+x}{1+x^2}\right)=$$
$$=\frac12\left(1+x+x^2+\ldots+1-x^2+x^4-\ldots+x-x^3+x^5-\ldots\right)=$$
$$\frac12\sum_{k=0}^\infty \left(x^k+(-1)^{\left\lfloor \frac k2\right\rfloor}x^k\right)=\frac12\sum_{k=0}^\infty\left(1+(-1)^{\left\lfloor \frac k2\right\rfloor}\right)x^k\implies$$
so the coefficient of $\;x^n\;$ is
$$\frac12\left(1+(-1)^{\left\lfloor \frac n2\right\rfloor}\right)$$
observe this coefficient vanishes whenever
$$(-1)^{\left\lfloor \frac n2\right\rfloor}=-1\iff \left\lfloor \frac n2\right\rfloor\;\;\text{is odd}\iff n=2,3\pmod 4$$
Another way: since $\;x^4-1=(x^2-1)(x^2+1)=(x+1)(x^3-x^2+x-1)\;$ , we have
$$\frac1{1-x+x^2-x^3=}=\frac{1+x}{1-x^4}=\left(1+x\right)\left(1+x^4+x^8+\ldots\right)=$$
$$=\sum_{k=0}^\infty\left(x^{4k}+x^{4k+1}\right)=\sum_{k=0}^\infty\left(1+x\right)x^{4k}$$
Both ways above render the same answer (check this), so I'd say definitely your book is wrong...or else you're looking at the wrong question's answer...or else I am wrong, of course. Twice.