Question

If $i^2=-1$, then $1+i^2+i^4+i^6+i^8+.............to(2n+1)$ terms is equal to

• A. $0$
• B. $1$
• C. $3i$
• D. $4i$

Answer 1

Answer: (B)

$1$

Solution:- (B) +1

As the given sequence is in G.P.,

Sum of N terms of G.P. with $r<1,S_N$ = $\frac{a(1-r^N)}{(1-r)}$

here,

a = first term = 1

r = common ratio $=\frac{a_2}{a_1}=\frac{i^2}{1}=i^2=-1\{\because i^2=-1\}$

As the total no. of terms are $(2n+1)$

$\Rightarrow N=2n+1$

$\therefore S_{2n+1}=\frac{1\times (1-{(-1)}^{2n+1})}{1-(-1)}$

$\Rightarrow {(-1)}^{2n+1}=-1\{\because 2n+1=odd\}$

$\therefore S_{2n+1}=\frac{(1-(-1))}{(1-(-1))}=\frac{2}{2}=1$