Question

Simplify $i^2+i^4+i^6+...+(2n+1)$ terms.

• A. $i$
• B. $-i$
• C. $1$
• D. $-1$

Answer 1

Answer: (D)

$-1$

Here we given a series with complex numbers. It is G.P series

$a=i^2$ & $r=i^2$

So, the $(2n+1)$th term will be

$a.r^{2n+1)-1=a.r^{2n}=i^2(i^2{)}^{2}n}$

$=i^{2+4n}$

Here, the series $i^2+i^4+i^6+i^8.....+i^{2+4n}$

make a pair of zero as $i^2=-14i^4=1$

$S_n=(i^2+i^4)+(i^6+i^8)+....+(i^{4n-2}+i^{4n})+i^{2+4n}$

(as terms is even)

$S_n=0+0+.....+0+i^{2+4n}$

As we know, $i^2=-1$ & $i^4=1$

So, $S_n=i^2\cdot i^{4n}$

$=(-1)\cdot (1{)}^n$

$S_n=-1$

So, $i^2+i^4+i^6+.....+(2n+1)$ terms

$=-1$.