Question

Suppose $n$ is a natural number such that $|i+2i^2+3i^3+...+n{i}^n|=18\sqrt{2}$
where $i$ is the square root of $-1$. Then $n$ is.

• A. $9$
• B. $18$
• C. $36$
• D. $72$

Answer 1

Answer: (C)

$36$

$S=i^1+{2i}^2+3i^3+.....n{i}^{n}iS=i^2+{2i}^3+3i^4.....n{i}^{n+1}S-iS=i^1+i^2+i^3+.....i^n-n{i}^{n+1}S(1-i)=\frac{i(1-i^n)}{1-i}-n{i}^{n+1}S=\frac{i(1-i^n)}{{(1-i)}^2}-\frac{n{i}^{n+1}}{(1-i)}S=\frac{-(1-i^n)}{2}-\frac{n{i}^{n+1}}{(1-i)}{z}_1=\frac{-(1-i^n)}{2},z_2=\frac{n{i}^{n+1}}{(1-i)}\left|z_1\right|=\frac{1}{\sqrt{2}}or0,\left|z_2\right|=\frac{n}{\sqrt{2}}\therefore \frac{n}{\sqrt{2}}=18\sqrt{2}n=36$

Hence, option $C$ is correct.