If {i^2} = -1,then the sum i + {i^2} + {i^3} | vedclass

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

Question

(D) The given series is a geometric progression with first term $a = i$ and common ratio $r = i$.The sum of the first $n$ terms of a geometric progres

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) The given series is a geometric progression with first term $a = i$ and common ratio $r = i$.The sum of the first $n$ terms of a geometric progression is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.For $n = 1000$ terms,$S_{1000} = \frac{i(1 - i^{1000})}{1 - i}$.Since $i^4 = 1$,we have $i^{1000} = (i^4)^{250} = 1^{250} = 1$.Substituting this into the sum formula: $S_{1000} = \frac{i(1 - 1)}{1 - i} = \frac{i(0)}{1 - i} = 0$.

If ${i^2} = -1$,then the sum $i + {i^2} + {i^3} + \dots$ up to $1000$ terms is equal to

Similar Questions

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

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

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

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

Express the given complex number in the form $a+ib$: $(1-i)-(-1+i6)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module