Let $a_{n} = i^{(n + 1)^{2}}$, where $i = \sqrt{- 1}$ and $n = 1,\ 2,\ 3,\ldots$Then the value of $a_{1} + a_{3} + a_{5} + ... + a_{25}$ is

Question

Dr. Abubacker · Accepted Answer

(a)
Given, $a_{n} = i^{(n + 1)^{2}}$
Here, $a_{1} = i^{2^{2}} = 1,\ a_{2} = i^{3^{2}} = i$,
$a_{3} = i^{4^{2}} = 1,\ \ a_{4} = i^{5^{2}} = i,$
$a_{5} = i^{6^{2}} = 1,\ldots$
Here, we see that for all odd values of $n$, we get the value of $a_{n}$ is 1
$	herefore\ \ \ a_{1} + a_{3} + a_{5} + \ldots + a_{25} =$
$$

Q. Let $a_{n} = i^{(n + 1)^{2}}$, where $i = \sqrt{- 1}$ and $n = 1,\ 2,\ 3,\ldots$Then the value of $a_{1} + a_{3} + a_{5} + ... + a_{25}$ is

Explanation

Reviewed by Dr. Abubacker

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