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UPSC NDA Math Model Paper 8

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Question : 49 of 120

Marks: +1, -0

What is the value of

1+i^{2}+i^{4}+i^{6}+\dots+i^{100}

i=\sqrt{1} ?

0
1
-1
None of the above

Solution:

Consider

1+i^{2}+i^{4}+i^{6}+\dots+i^{100}

Here,

i=\sqrt{1}

\Rightarrow i^{2}=-1

Series becomes,

⇒ 1 - 1 + 1 - 1 + .... + 1

which is GP with common ratio r = -1 and first term a = 1.

\Rightarrow 1-1+1-1+\dots+1=\frac{a(1-r^{n})}{1-r}

Here

n=51

\Rightarrow 1-1+1-1+\dots+1=\frac{1[1-(-1)^{51}]}{1-(-1)}

\Rightarrow 1-1+1-1+\dots+1=1

Hence,

1+i^{2}+i^{4}+i^{6}+\dots+i^{100}=1

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