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

The correct option is D

None of these

Explanation for the correct option:

Given expression, $i^{\left(1+3+5+.....+(2n+1)\right)}$

It is clear that the terms in the power of $i$ are in A.P. with starting term as $1$ and common difference as $2$ and $n+1$ number of terms.

So,

$$\begin{gathered} i^{\left(1+3+5+.....+(2n+1)\right)}=i^{\frac{n+1}{2}\left[2\left(1\right)+(n+1-1)2\right]} \\ \Rightarrow i^{\left(1+3+5+.....+(2n+1)\right)}=i^{\frac{n+1}{2}\left[2+2n\right]} \\ \Rightarrow i^{\left(1+3+5+.....+(2n+1)\right)}=i^{{\left[1+n\right]}^2} \end{gathered}$${Since, the formula of the sum of $n$ terms of an A.P is $\frac{n}{2}(2a+(n-1\left)d\right)$}

Since, when $n$ is even then ${\left(n+1\right)}^2$ is odd.

So, The value of the given expression $i^{\left(1+3+5+.....+(2n+1)\right)}$ is $1$.

Since, when $n$ is odd then ${\left(n+1\right)}^2$ is even.

So, The value of the given expression $i^{\left(1+3+5+.....+(2n+1)\right)}$ is $i$.

Therefore, the value of the given expression is $1$ when $n$ is even and $i$ when $n$ is odd.

Hence, option D is the correct answer.