Complex Numbers 1 Question 8
8.
The value of $\operatorname{sum} \sum _{n=1}^{13}\left(i^{n}+i^{n+1}\right)$, where $i=\sqrt{-1}$, equals
(a) $i$
(b) $i-1$
(c) $-i$
(d) 0
(1998, 2M)
Show Answer
Answer:
Correct Answer: 8. (b)
Solution:
- $\sum _{n=1}^{13}\left(i^{n}+i^{n+1}\right)=\sum _{n=1}^{13} i^{n}(1+i)=(1+i) \sum _{n=1}^{13} i^{n}$
$ \begin{aligned} & =(1+i)\left(i+i^{2}+i^{3}+\ldots+i^{13}\right)=(1+i) \quad (\frac{i-\left(1-i^{13}\right)}{1-i}) \\ & =(1+i) (\frac{i(1-i)}{1-i})=(1+i) i=i-1 \end{aligned} $
Alternate Solution
Since, sum of any four consecutive powers of iota is zero.
$\therefore \sum _{n=1}^{13}\left(i^{n}+i^{n+1}\right) $
$ =\left(i+i^{2}+\ldots+i^{13}\right) + \left(i^{2}+i^{3}+\ldots+i^{14}\right)$
$=i+i^{2}=i-1$
