Q69 / 180 · COMEDK 2018 · Mathematics

The maximum value of $n < 101$, such that $1+\sum_{k-1}^{n} i^{k}=0$ is

  • A
    96
  • B
    97
  • Correct answer
    C
    99
  • D
    100

Solution

We have, $\sum_{n=1}^{n} i^{k}=i+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+\ldots$ Now as, $i^{4}=i^{8}=i^{12}=\ldots i^{n}=1$ and $i+i^{2}+i^{3}+i^{4}=0$ So, $\sum_{n=1}^{n} i^{k}=i+i^{2}+i^{3}+i^{4}+i+i^{6}+\ldots i^{7}+i^{8}+i^{2}+\ldots$ $$ \begin{aligned} &=i+i^{2}+i^{3}+i^{4}+\left(i+i^{2}+i^{3}+i^{4}\right)+\left(i+i^{2}+i^{3}+i^{4}\right)+\ldots \\ &=0+0+0+\ldots \end{aligned} $$ When we take $n=99$ after 96 th term, we will be left with $0+i+i^{2}+i^{3}=-1$ $\mathrm{So}$, for $$ \Rightarrow \quad 1+\sum_{k=1}^{n} i^{k}=0 $$

Hence, the correct answer is option: 'C'.

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