Question

State true or false:

If $\omega \ne 1$ and $\omega$ is a nth root of unity then the value of $1+4\omega +9{\omega }^2+16{\omega }^3\cdots +n^2{\omega }^{n-1}$ is $\frac{1}{{(\omega -1)}^2}[n^2\omega -n(n+2)]$.

Type 1 for true and 0 for false

Answer 1

$1+x+x^2+\cdots +x^n=\frac{x^{n+1}-1}{x-1}$ G.P.Differentiate both sides w.r. to x, we get
$1+2x+3x^2+\cdots +n{x}^{n-1}$
$=\frac{(n+1)x^n}{x-1}-\frac{x^{n+1}-1}{{(x-1)}^2}$ ...(1)
We have used product formula instead of quotient formula for differentiation keeping in view the form of the question we multiply both sides of (1) by x and then we will differentiate once again by using product formula instead of quotient formula
$x+2x^2+3x^2+\cdots +n{x}^n=\frac{(n+1)x^{{}^{n+1}}}{x-1}-\frac{x^{n+2}-x}{{(x-1)}^2}$
Differentiate both sides w.r. to x, we get
$1+4x+9x^2+\cdots n^2{x}^n=\frac{{(n+1)}^2{x}^n}{x-1}-\frac{(2n+3)x^{n+1}-1}{{(x-1)}^2}+\frac{2(x^{n+2}-x)}{{(x-1)}^2}$
or $1+2^{2}x+3^2.x^2+\cdots +n^2{x}^n=\frac{{(n+1)}^2{x}^n}{x-1}-\frac{(2n+3)x^{n+1}-1}{{(x-1)}^2}+\frac{2(x^{n+2}-x)}{{(x-1)}^3}$
Now put $x=\omega$ and ${\omega }^n=1$ in both sides and simplify R.H.S.
$\therefore L.H.S.=\frac{{(n+1)}^2}{\omega -1}-\frac{(2n+3)\omega -1}{{(\omega -1)}^2}+\frac{2\omega (\omega -1)}{{(\omega -1)}^3}$
$=\frac{1}{{(\omega -1)}^2}[{(n+1)}^2(\omega -1)-(2n+3)\omega +1+2\omega ]$
$=\frac{1}{{(w-1)}^2}[\omega \{(n^2+2n+1)-2n-3+2\}-(n^2+2n+1)+1]$
$=\frac{1}{{(\omega -1)}^2}[n^2\omega -n(n+2)]$