If {i^2} = -1,then the value of sum_{n=1}^{20 | vedclass

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

Question

(C) The given expression is a geometric series: $\sum_{n=1}^{200} {i^n} = i + i^2 + i^3 + \dots + i^{200}$.Here,the first term $a = i$,the common rati

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) The given expression is a geometric series: $\sum_{n=1}^{200} {i^n} = i + i^2 + i^3 + \dots + i^{200}$.Here,the first term $a = i$,the common ratio $r = i$,and the number of terms $n = 200$.The sum of a geometric series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.Substituting the values: $S_{200} = \frac{i(1 - i^{200})}{1 - i}$.Since $i^4 = 1$,we have $i^{200} = (i^4)^{50} = 1^{50} = 1$.Therefore,$S_{200} = \frac{i(1 - 1)}{1 - i} = \frac{0}{1 - i} = 0$.

If ${i^2} = -1$,then the value of $\sum_{n=1}^{200} {i^n}$ is

Similar Questions

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

Let ${\left( { - 2 - \frac{1}{3}i} \right)^3} = \frac{{x + iy}}{{27}}$ where $i = \sqrt{-1}$ and $x, y$ are real numbers,then $y - x$ equals

$\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right) \left( \frac{3 + 4i}{2 - 4i} \right) = $

Express the given complex number in the form $a+ib$: $(5i)\left(-\frac{3}{5}i\right)$

If $4x + i(3x - y) = 3 + i(-6)$,where $x$ and $y$ are real numbers,then find the values of $x$ and $y$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module