Complex Numbers And Quadratic Equations question 395
Question: If $ i^{2}=-1 $ , then the value of $ \sum\limits_{n=1}^{200}{i^{n}} $ is [MP PET 1996]
Options:
A) $ 50 $
B) - 50
C) 0
D) 100
Show Answer
Answer:
Correct Answer: C
Solution:
$ \sum\limits_{n=1}^{200}{i^{n}=i+i^{2}+i^{3}+….+i^{200}=\frac{i(1-i^{200})}{1-i}} $ (since G.P.) $ =\frac{i(1-1)}{1-i}=0 $ .
BETA
