Question

If $\omega \ne 1,and\omega$is a n${}^{th}$ root of unity, then the value of 2 + 4$\omega +9{\omega }^2+16{\omega }^3+\dots \dots \dots ..+n^2{\omega }^{n-1}is:$

• A. -nw
• B.
• C.
• D. None of these

Answer 1

The correct option is C

We have, for x $\ne$ 1,
$1+x+x^2+x^3+\dots \dots \dots \dots .+x^n=\frac{(x^{n+1}-1)}{(x-1)}$
Differentiating w.r.t x, we get,
$1+2x+3x^2+\dots \dots \dots \dots .+{nx}^{n-1}=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{{(x-1)}^2}$
Multiplying both side by x, we get
$x+2x^2+3x^3+\dots \dots \dots \dots \dots +{nx}^n=\frac{(n+1)x^{n+1}}{x-1}-\frac{x^{n+2}-1}{{(x-1)}^2}$
Differentiating again w.r.t x, we get
$1+2^{2}x+3^2{x}^2+\dots \dots \dots .+n^2{x}^{n-1}=$ $\frac{{(n+1)}^2{x}^n}{x-1}-\frac{(2n+3)x^{n+1}}{{(x-1)}^2}+\frac{2(x^{n+2}-x)}{{(x-1)}^3}$
Putting $x=\omega$and using ${\omega }^n=1,weget$
$1+4\omega +9{\omega }^2+\dots \dots ..+n^2{\omega }^{n-1}$
$=\frac{{(n+1)}^2}{\omega -1}-\frac{(2n+3)\omega -1}{{(\omega -1)}^2}+\frac{2({\omega }^2-\omega )}{{(\omega -1)}^3}$
$=\frac{{(n+1)}^2(\omega -1)-(2n+3)\omega +1+2\omega }{{(\omega -1)}^2}$
$=\frac{\{{(n+1)}^2-(2n+3)+2\}\omega -(n^2+1)+1}{{(\omega -1)}^2}$
$=\frac{n^2\omega -n^2-2n}{{(\omega -1)}^2}=\frac{n^2(\omega -1)-2n}{{(\omega -1)}^2}$