1.

What is the value of \(1+i^2 + i^4 + i^6 + ...+ i^{100}\) where \(i =\sqrt{1} \ ?\)

Answer» Correct Answer - Option 2 : 1

Concept:

Let \(\rm a_1 + a_2 + a_3 + .... + a_n \) be geometric progression series then sn =  \(\rm \sum a_n = \dfrac {a_1(1 - r^n)}{1-r}\), Where r is common ratio

Calculations:

Consider 

\(1+i^2 + i^4 + i^6 + ...+ i^{100} \)

Here,  \(i =\sqrt{1} \)

⇒ \(\rm i^2 = -1\)

Series becomes,

⇒ 1 - 1 + 1 - 1 + .... + 1

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

⇒ 1 - 1 + 1 - 1 + .... + 1 = \(\rm \dfrac {a (1 - r^n)}{1- r}\)

Here n = 51.

⇒ 1 - 1 + 1 - 1 + .... + 1 = \(\rm \dfrac {1 [1 -(-1)^{51}]}{1- (-1)}\)

⇒ 1 - 1 + 1 - 1 + .... + 1 = 1

Hence,  \(1+i^2 + i^4 + i^6 + ...+ i^{100}\) = 1


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