If an-in-12- where i - -1 and n - 1- 2- 3- Then - the value of a1 - a3 - a5 - a25 is

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Adjoint of a Matrix

If an=in+12, ...

Question

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

$13$

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$13 + i$

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$13 - i$

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$12$

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$12 - i$

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Solution

The correct option is A
$13$

Explanation for the correct option:
Find the value of $a_{1} + a_{3} + a_{5} + . . . + a_{25}$ :
Given, $a_{n} = i^{{(n + 1)}^{2}}$
$a_{1} + a_{3} + a_{5} + . . . + a_{25} = i^{{(1 + 1)}^{2}} + i^{{(3 + 1)}^{2}} + i^{{(5 + 1)}^{2}} + . . . + i^{{(25 + 1)}^{2}}$
As we know, There are $13$ odd numbers from $1$ to $25$ .
$= i^{4} + i^{16} + i^{36} + \dots i^{676} = 1 + 1 + 1 + \dots 1 = 13$ $[∵ i^{4} = 1]$
Hence, Option ‘A’ is Correct.

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If an=i(n+1)2, where i=√-1 and n=1,2,3... Then , the value of a1+a3+a5+...+a25 is

The correct option is A 13Explanation for the correct option:Find the value of a1+a3+a5+...+a25:Given, an=i(n+1)2a1+a3+a5+...+a25=i1+12+i3+12+i5+12+...+i25+12As we know, There are 13 odd numbers from 1 to 25.=i4+i16+i36+…i676=1+1+1+…1=13 ∵i4=1Hence, Option ‘A’ is Correct.

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