i^2+i^3+ldots+i^{4000}= | vedclass

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

Question

(D) We know that the sum of four consecutive powers of $i$ is zero,i.e.,$i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$ for any integer $n$. The given series is $S =

Vedclass · Accepted Answer

$-i$

Vedclass · Answer

(D) We know that the sum of four consecutive powers of $i$ is zero,i.e.,$i^n+i^{n+1}+i^{n+2}+i^{n+3}=0$ for any integer $n$. The given series is $S = i^2+i^3+i^4+\ldots+i^{4000}$. This series contains $4000-2+1 = 3999$ terms. We can group these terms into sets of four. Since $3999 = 4 	imes 999 + 3$,there are $999$ groups of four consecutive powers of $i$ (each summing to $0$) and $3$ remaining terms. The sum of the first $3996$ terms is $0$. Thus,$S = i^{3997} + i^{3998} + i^{3999} + i^{4000} = 0$ is not the correct approach here; rather,we observe the sequence $i^2, i^3, i^4, \ldots, i^{4000}$. The sum is $\sum_{k=2}^{4000} i^k = \frac{i^2(1-i^{3999})}{1-i} = \frac{-1(1-i^3)}{1-i} = \frac{-1(1-(-i))}{1-i} = \frac{-(1+i)}{1-i} = \frac{-(1+i)(1+i)}{(1-i)(1+i)} = \frac{-(1+2i+i^2)}{1-i^2} = \frac{-(1+2i-1)}{1+1} = \frac{-2i}{2} = -i$.

$i^2+i^3+\ldots+i^{4000}=$

Similar Questions

If $i=\sqrt{-1}$,then $(1+i)^{10}+(1-i)^{10}=$

If $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$,then $x$ is equal to

Express the following expression in the form of $a+ib$:
$\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}$

The value of $\frac{i^{248}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{249}+i^{247}+i^{245}+i^{243}+i^{241}}$,where $i=\sqrt{-1}$,is

$\sum_{k=0}^{40} i^k = x + iy \Rightarrow x^{100} + x^{99}y + x^{242}y^2 + x^{97}y^3 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module