Question

The value of $\mathrm{i}^{(1+3+5 . .(2 n+1)}$
 is
A .if $n$ is even, $-i$ if $n$ is odd
B. $i$ if $n$ is even, $-1$ if $n$ is odd
C. $i$if $n$ is odd, $-1$ if $n$ is even
D. none of these

Answer 1

Hint: In this question, first, we find the sum of (n+1) terms by using the sum formula. Then we put the values of n=1,2,3, 4, 5, and then the values which we find out by putting the different values of n are put in the powers of i to find out the correct option.

Formula used:
$S_n=\dfrac{n}{2}[2a+\lgroup~n-1\rgroup~d]$

Complete step-by-step solution:
Given that
$\mathrm{i}^{(1+3+5 . .(2 n+1)}$
$1+3+5+\ldots(2 n+1)=$ sum of $(n+1)$ terms of A.P
Sum to $(n+1)$terms is
We know ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Here $a=1,d=3-1=2$
Then ${{S}_{n+1}}=\dfrac{n+1}{2}\left[ 2(1)+\left( n+1-1 \right)(2) \right]$
Then ${{S}_{n+1}}=\dfrac{n+1}{2}\left[ 2+2n \right]$
And ${{S}_{n+1}}=\dfrac{n+1}{2}\times 2\left[ 1+n \right]$
${{S}_{n+1}}=(n+1)(1+n)$
Hence ${{S}_{n+1}}={{(n+1)}^{2}}$
This implies, $\mathrm{i}^{(1+3+5 . .(2 n+1)}=\mathrm{i}^{{(n+1)}^{2}}$..........(i)
Now we are substituting values to the resultant equation and checking whether it is odd or even
 Put $n=1$
$S_{2}=(2+1)^{2}=3^{2}=9$
 $S_{3}=(3+1)^{2}=16$
 $S_{4}=(4+1)^{2}=25$
 $S_{5}=(5+1)^{2}=36$
Now substitute the sum values to the power of $\mathrm{i}$ in each condition
When $n=1,{{i}^{9}}={{({{i}^{2}})}^{4}}i={{(-1)}^{4}}i=i$
$n=2,{{i}^{16}}={{({{i}^{2}})}^{8}}={{(-1)}^{8}}=1$
$n=3,{{i}^{25}}={{({{i}^{2}})}^{12}}i={{(-1)}^{12}}i=i$
$n=4,{{i}^{36}}={{({{i}^{2}})}^{18}}={{(-1)}^{18}}=1$
 As a result, when$\mathrm{n}$ is even, the value is 1, and when $\mathrm{n}$ is odd, it is $\mathrm{i}$
There is no option that matches our answer.

So the correct answer is option D.

Note: The difference between any two consecutive integers in an arithmetic progression (AP) sequence of numbers is always the same amount. For instance, the natural number sequence 1, 2, 3, 4, 5, 6,... is an example of an arithmetic progression. It has a common difference of 1 between two succeeding terms (let's say 1 and 2). (2 -1). We can see that the common difference between two subsequent terms will be equal to 2 in both the case of odd and even numbers.