sum_{k=0}^{40} i^k = x + iy Rightarrow x^{100 | vedclass

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(D) Given the sum $\sum_{k=0}^{40} i^k = x + iy$. Since the sum of four consecutive powers of $i$ is zero,i.e.,$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given the sum $\sum_{k=0}^{40} i^k = x + iy$. Since the sum of four consecutive powers of $i$ is zero,i.e.,$i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0$,we can group the terms. The sum has $41$ terms from $k=0$ to $k=40$. $\sum_{k=0}^{40} i^k = i^0 + (i^1 + i^2 + i^3 + i^4) + \dots + (i^{37} + i^{38} + i^{39} + i^{40}) = 1 + 0 + \dots + 0 = 1$. Thus,$x + iy = 1 + 0i$,which gives $x = 1$ and $y = 0$. Substituting these values into the expression: $x^{100} + x^{99}y + x^{242}y^2 + x^{97}y^3 = (1)^{100} + (1)^{99}(0) + (1)^{242}(0)^2 + (1)^{97}(0)^3 = 1 + 0 + 0 + 0 = 1$.

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

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