Question

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

• A. i if n is even, - i if n is odd
• B. 1 if n is even, - 1 if n is odd
• C. 1 if n is odd, - 1 if n is even
• D. i if n is even, - 1 if n is odd

Answer 1

The correct option is C

1 if n is odd, - 1 if n is even

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

Clearly series is A.P with common difference = 2

∵ $T_n$ = 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)2]

$s_{n+1}=\frac{n+1}{2}[2+2n]=(n+1{)}^{2}i.e.(i^{n+1}{)}^2$

Now put n = 1,2,3,4,5,........

n = 1, z = $i^4$ = 1, n = 2, z = $i^5$ = -1 ,

n = 3, z = $i^8$ = 1, n = 4, z = $i^{10}$ = -1 ,

n = 5, z = $i^{12}$ = 1, .........