Question

The value of the sum ${\sum }_{n=1}^{13}(i^n+i^{n+1})$ where $i=\sqrt{-1}$, is

• A. $1+i$
• B. $i-1$
• C. $-i$
• D. $0$

Answer 1

The correct option is A $1+i$

Let $L=\sum _{n=1}^{13}(i^n+i^{n-1})$

$L=\sum _{n=1}^{13}{i}^n+\sum _{n=1}^{13}{i}^{n-1}$

$=(i^1+i^2+i^3+i^4+i^5+i^6+i^7+i^8+i^9+i^{10}+i^{11}+i^{12}+i^{13})$

$+(i^0+i^1+i^2+i^3+i^4+i^5+i^6+i^7+i^8+i^9+i^{10}+i^{11}+i^{12})$

$=i^0+2(i^1+i^2+i^3+i^4+i^5+i^6+i^7+i^8+i^9+i^{10}+i^{11}+i^{12})+i^{13}$

We have $i^1+i^2+i^3+i^4=i-1-i+1=0$

Using this, we have

$=i^0+2(i+i^2+i^3+i^4)+2i^4(i+i^2+i^3+i^4)+2i^8(i+i^2+i^3+i^4)+i^{13}$

$=1+0+0+0+{\left(i^4\right)}^3\times i$

$L=1+i$ since $i^4=1$

$\therefore \sum _{n=1}^{13}(i^n+i^{n-1})=1+i$