If sumlimits_{k = 0}^{100} {{i^k}} = x + iy,t | vedclass

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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

$x = 1, y = 0$

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(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:

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