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

Let $z=i^{[1+3+5+\mathrm{....}+(2n+1\left)\right]}$ 

Clearly series is A.P. with common difference = 2

$\because T_n=2n-1$ and $T_{n+1}=2n+1$

So, number of terms in A. P. $=n+1$

Now, $S_{n+1}=\frac{n+1}{2}[2.1+(n+1-1\left)2\right]$

$\Rightarrow S_{n+1}=\frac{n+1}{2}[2+2n]={(n+1)}^2$ 

                 Z= $i^{{(n+1)}^2}$

Now put $n=1,2,3,4,5,\mathrm{.....}$

$n=1,z=i^4=1$

 $n=2,z=i^9=i$

$n=3,z=i^{16}=1,$

 $n=4,z=i^{25}=i$

$n=5,z=i^{36}=1,\mathrm{........}$