Suppose n is a natural number such that |i + | vedclass

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(C) Let $S_n = i + 2i^2 + 3i^3 + \ldots + ni^n$ $(i)$Multiplying by $i$,we get $iS_n = i^2 + 2i^3 + \ldots + (n-1)i^n + ni^{n+1}$ (ii)Subtracting (ii)

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$36$

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(C) Let $S_n = i + 2i^2 + 3i^3 + \ldots + ni^n$ $(i)$Multiplying by $i$,we get $iS_n = i^2 + 2i^3 + \ldots + (n-1)i^n + ni^{n+1}$ (ii)Subtracting (ii) from $(i)$:$S_n(1 - i) = i + i^2 + i^3 + \ldots + i^n - ni^{n+1}$$S_n(1 - i) = \frac{i(1 - i^n)}{1 - i} - ni^{n+1}$Since $(1 - i)^2 = 1 - 2i + i^2 = -2i$,we have:$S_n = \frac{i(1 - i^n)}{(1 - i)^2} - \frac{ni^{n+1}}{1 - i} = \frac{i(1 - i^n)}{-2i} - \frac{ni^{n+1}(1 + i)}{2} = \frac{i^n - 1}{2} - \frac{ni^{n+1}(1 + i)}{2}$For large $n$,the term involving $n$ dominates the magnitude. Given $|S_n| = 18\sqrt{2}$,we consider the magnitude of the dominant term:$|S_n| \approx |\frac{ni^{n+1}(1 + i)}{2}| = \frac{n|i^{n+1}| |1 + i|}{2} = \frac{n \cdot 1 \cdot \sqrt{2}}{2} = \frac{n}{\sqrt{2}}$Setting $\frac{n}{\sqrt{2}} = 18\sqrt{2}$,we get $n = 18 	imes 2 = 36$.

Suppose $n$ is a natural number such that $|i + 2i^2 + 3i^3 + \ldots + ni^n| = 18\sqrt{2}$,where $i = \sqrt{-1}$. Then,$n$ is

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