Question

The value of $i^{1+3+5+....+\left(2n+1\right)}$ is _______.

Answer 1

Consider the value $i^{1+3+5+\mathrm{7...........}(2n+1)}$

Let, $x=$ $1+3+5+\mathrm{7......}(2n+1)$,then $i^{1+3+5+\mathrm{7...........}(2n+1)}=i^x$

…….(1)

This is A.P. series which have first term is a=1, common difference is (d)=2 and number of term is $N=2n+1$

Now sum of A.P. is, $\because s_n=\frac{n}{2}[2a+(n-1)d]$

Thus ,

$x$ $=\frac{2n+1}{2}[2(2n+1)+(2n\left)2\right]$

$=\frac{2n+1}{2}[(2n+1)+(2n\left)\right]$

$=\frac{2n+1}{2}\left[(4n+1)\right]$

$x$ $=\frac{(2n+1)(4n+1)}{2}$

Put the value of x in equation 1^st

Hence,$i^{1+3+5+\mathrm{7...........}(2n+1)}=i^{\frac{(2n+1)(4n+1)}{2}}$