Sequence And Series Question 382
Question: The sum of $ i-2-3i+4+……… $ upto 100 terms, where $ i=\sqrt{-1} $ is
Options:
A) $ 50(1-i) $
B) $ 25i $
C) $ 25(1+i) $
D) $ 100(1-i) $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ S=i-2-3i+4+5i+…..+100i^{100} $
$ \Rightarrow $ $ S=i+2i^{2}+3i^{3}+4i^{4}+5i^{5}+……..+100i^{100} $
Þ $ iS=,i^{2}+2i^{3}+3i^{4}+4i^{5}+……..+99i^{100}+100i^{101} $
$ \therefore $ $ S-iS=[i+i^{2}+i^{3}+i^{4}+……+i^{100}]-100i^{101} $
$ \Rightarrow $ $ S(1-i)=0-100i^{101}=-100i $
$ \therefore $ $ S=\frac{-100i}{1-i}=-50i(1+i)=-50(i-1)=50(1-i) $ .
BETA
