Complex Numbers And Quadratic Equations question 398
Question: The value of $ {i^{1+3+5+…+(2n+1)}} $ is [AMU 1999]
Options:
A) i if n is even, - i if n is odd
B) 1 if n is even, - 1 if n is odd
C) 1 if n is odd, - 1 if n is even
D) i if n is even, - 1 if n is odd
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ z={i^{[1+3+5+….+(2n+1)]}} $ Clearly series is A.P. with common difference = 2 $ \because T_{n}=2n-1 $ and $ {T_{n+1}}=2n+1 $ So, number of terms in A. P. $ =n+1 $ Now, $ {S_{n+1}}=\frac{n+1}{2}[2.1+(n+1-1)2] $
$ \Rightarrow {S_{n+1}}=\frac{n+1}{2}[2+2n]={{(n+1)}^{2}} $ i.e. $ {i^{{{(n+1)}^{2}}}} $ Now put $ n=1,2,3,4,5,….. $ $ n=1,z=i^{4}=1 $ , $ n=2,z=i^{6}=-1 $ , $ n=3,z=i^{8}=1 $ , $ n=4,z=i^{10}=-1 $ , $ n=5,z=i^{12}=1,…….. $
BETA
