Question

The value of $i^{1+3+5+.....+(2n+1)}$ is________.

Answer 1

Solution:-

$i^{1+3+5+.....+(2n+1)}$

Let $1+3+5+.....+(2n+1)=S$

Total no. of terms $=(n+1)$

$\therefore 2S=(1+3+5+...+(2n+1))+((2n+1)+(2n-1)+(2n-3)+...+1)$

$\Rightarrow 2S=[(2n+2)+(2n+2)+(2n+2)+...+(2n+2)]$

$\Rightarrow 2S=(2n+2)\times (n+1)$

$\Rightarrow 2S=2{(n+1)}^2$

$\Rightarrow S={(n+1)}^2$

$\therefore i^{1+3+5+.....+(2n+1)}$

$=i^{{(n+1)}^2}$

$=i^{n^2+2n+1}$

$=i^{n^2}.i^{2n}.i^1$

$=i^{n^2}.(-1).i[\because i^{n+1}=i,i^{n+2}=-1,i^{n+3}=-i,i^{n+4}=1]$

$=-i^{n^2+1}$

Hence, $i^{1+3+5+.....+(2n+1)}=-i^{n^2+1}$.