Question

Find $a$, $b$, $c$ so that the coefficient of $x^n$ in the expansion of $\frac{a+bx+c{x}^2}{{(1-x)}^3}$ may be $n^2+1$.

Answer 1

Let's start this by putting $n=0,1,2,3,........$

Successively we see that the required expansion will have coefficients $1,2,5,\mathrm{10........}$

i.e. $\frac{a+bx+c{x}^2}{{(1-x)}^3}=(1+2x+5x^2+10x^3+.....)$

$\Rightarrow a+bx+c{x}^2={(1-x)}^3(1+2x+5x^2+10x^3+.....)$

By equating coefficients, we get

$a=1,b=-3+2=-1$

$c=3-6+5=2$

Hence, $a=1,b=-1$ and $c=2$