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