sumlimits_{n = 1}^{50} {{i^{2n-1}}} is equal | vedclass

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

Question

(A) The given sum is $S = \sum\limits_{n = 1}^{50} {{i^{2n - 1}}} = i^1 + i^3 + i^5 + i^7 + \dots + i^{99}$.This is a geometric progression with first

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The given sum is $S = \sum\limits_{n = 1}^{50} {{i^{2n - 1}}} = i^1 + i^3 + i^5 + i^7 + \dots + i^{99}$.This is a geometric progression with first term $a = i$,common ratio $r = i^2 = -1$,and number of terms $n = 50$.The sum of a geometric progression is given by $S = \frac{a(1 - r^n)}{1 - r}$.Substituting the values: $S = \frac{i(1 - (-1)^{50})}{1 - (-1)} = \frac{i(1 - 1)}{2} = \frac{i(0)}{2} = 0$.

$\sum\limits_{n = 1}^{50} {{i^{2n-1}}}$ is equal to (where $i = \sqrt{-1}$)

Similar Questions

If $\frac{(1+i) x-i}{2+i}+\frac{(1+2 i) y+i}{2-i}=1$,then $(x, y)$ is equal to

If $i = \sqrt{-1}$ and $n$ is a positive integer, then $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ is equal to

Express the following in the form $a+ib$:
$\frac{5+\sqrt{2}i}{1-\sqrt{2}i}$

$(1 + i)^{10}$,where $i^2 = -1$,is equal to

The value of $\{i^{22}-(\frac{1}{i})^{35}\}^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module