If sumlimits_{k = 0}^{100} {{i^k}} = x + iy,t | 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: $\sum\limits_{k = 0}^{100} i^k = 1 + i + i^2 + \dots + i^{100}$.This is a geometric progression with f

Vedclass · Accepted Answer

$x = 1, y = 0$

Vedclass · Answer

(C) The given expression is a geometric series: $\sum\limits_{k = 0}^{100} i^k = 1 + i + i^2 + \dots + i^{100}$.This is a geometric progression with first term $a = 1$,common ratio $r = i$,and number of terms $n = 101$.The sum of a geometric series is given by $S_n = \frac{a(1 - r^n)}{1 - r}$.Substituting the values,we get $S_{101} = \frac{1(1 - i^{101})}{1 - i}$.Since $i^4 = 1$,we have $i^{101} = (i^4)^{25} 	imes i = 1^{25} 	imes i = i$.Therefore,$S_{101} = \frac{1 - i}{1 - i} = 1$.Thus,$1 + 0i = x + iy$.Equating the real and imaginary parts,we get $x = 1$ and $y = 0$.

If $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$,then the values of $x$ and $y$ are:

Similar Questions

$(1+i)^{2024}+(1-i)^{2024} = $

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 given complex number in the form $a+ib$: $i^{-39}$

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

By simplifying $i^{18}-3i^7+i^2(1+i^4)(i)^{22}$,we get

Vedclass Test Series

Exam Paper Generator

Online Exam Module