The value of the sum sumlimits_{n = 1}^{13} { | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We are given the sum $S = \sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $.This can be written as $S = \sum\limits_{n = 1}^{13} {i^n} + \sum\lim

Vedclass · Accepted Answer

$i - 1$

Vedclass · Answer

(B) We are given the sum $S = \sum\limits_{n = 1}^{13} {({i^n} + {i^{n + 1}})} $.This can be written as $S = \sum\limits_{n = 1}^{13} {i^n} + \sum\limits_{n = 1}^{13} {i^{n + 1}}$.Both are geometric series with $13$ terms.For the first series,the sum is $\frac{i(1 - i^{13})}{1 - i}$. Since $i^4 = 1$,$i^{13} = i^{12} 	imes i = 1 	imes i = i$.So,the first sum is $\frac{i(1 - i)}{1 - i} = i$.For the second series,the sum is $\frac{i^2(1 - i^{13})}{1 - i} = \frac{-1(1 - i)}{1 - i} = -1$.Therefore,$S = i + (-1) = i - 1$.

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

Similar Questions

If one root of the quadratic equation $ix^2 - 2(i + 1)x + (2 - i) = 0$ is $2 - i$,then the other root is

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$,where $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then $(a, b)$ is equal to

If $\left(\frac{1-i}{1+i}\right)^{96}=a+ib$,then $(a, b)$ is

If $z=(1-i)^3(x+i)$ is a purely imaginary number for $x=x_1$ and if $z$ is a purely real number for $x=x_2$,then $x_1 x_2=$

The real value of $\alpha$ for which $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real is

Vedclass Test Series

Exam Paper Generator

Online Exam Module