Search any question & find its solution
Question:
Answered & Verified by Expert
If $i^2=-1$, then the value of $\sum_{n=1}^{200} i^n$ is
Options:
Solution:
1404 Upvotes
Verified Answer
The correct answer is:
$0$
$\sum_{n=1}^{200} i^n=i+i^2+i^3+\ldots .+i^{200}=\frac{i\left(1-i^{200}\right)}{1-i}$(since G.P.)
$=\frac{i(1-1)}{1-i}=0$
$=\frac{i(1-1)}{1-i}=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.