Question:
Let $ {{a}_{n}}={{i}^{{{(n+1)}^{2}}}}, $ where $ i=\sqrt{-1} $ and $ n=1,2,3..... $ . Then the value of $ {{a}_{1}}+{{a}_{3}}+{{a}_{5}}+...+{{a}_{25}} $ is
Let $ {{a}_{n}}={{i}^{{{(n+1)}^{2}}}}, $ where $ i=\sqrt{-1} $ and $ n=1,2,3..... $ . Then the value of $ {{a}_{1}}+{{a}_{3}}+{{a}_{5}}+...+{{a}_{25}} $ is
Updated On: Jun 7, 2024
- 13
- $ 13+i $
- $ 13-i $
- 12
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The Correct Option is A
Solution and Explanation
Given, $ {{a}_{n}}={{i}^{{{(n+1)}^{2}}}} $
$ \therefore $ $ {{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,.......... $
$ \therefore $ For all odd values of n, we get the value of $ {{a}_{n}} $ is 1.
$ \therefore $ $ {{a}_{1}}+{{a}_{3}}+{{a}_{5}}+....+{{a}_{25}} $
$=\underbrace{1+1+1+......+1}_{13} $
$=13 $
$ \therefore $ $ {{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,.......... $
$ \therefore $ For all odd values of n, we get the value of $ {{a}_{n}} $ is 1.
$ \therefore $ $ {{a}_{1}}+{{a}_{3}}+{{a}_{5}}+....+{{a}_{25}} $
$=\underbrace{1+1+1+......+1}_{13} $
$=13 $
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Concepts Used:
Series
A collection of numbers that is presented as the sum of the numbers in a stated order is called a series. As an outcome, every two numbers in a series are separated by the addition (+) sign. The order of the elements in the series really doesn't matters. If a series demonstrates a finite sequence, it is said to be finite, and if it demonstrates an endless sequence, it is said to be infinite.
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The following are the two main types of series are: