Sequence And Series Question 331
Question: The sum of $ i-2-3i+4… $ up to 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:
[a] Let $ S=i-2-3i+4+5i….100 $ terms
$ \Rightarrow S=i+2i^{2}+3i^{3}+4i^{4}+5i^{5}….+100i^{100} $
$ \Rightarrow iS=i^{2}+2i^{3}+3i^{4}+99i^{100}+100i^{101} $
$ \Rightarrow S-iS=i+i^{2}+i^{3}+i^{4}+….+i^{100}-100i^{101} $
$ \Rightarrow S(1-i)=\frac{i(1-i^{100})}{1-i}-100i^{101} $
$ \Rightarrow S(1-i)=-100i $
$ \Rightarrow S=\frac{-100i}{1-i}=-50i(1+i)=-50(i-1) $ $ =50(1-i) $
BETA
