The sum of the first n terms of the series- 12-2-22-32-2-42-52-2-62- is n -2 n -12- when n is even- When n is odd- the sum is n2-2n-1B- -2n-12C- nn-1-2- n2-2n-1

The correct option is A When n is odd, 1^2 + 2.2^2 + 3^2 + 2.4^2 +…. + 2(n 1)^2 + n^2 = (n 1)n^2/2 + n^2, since (n 1) is even. = n^2/2 (n + 1) ...

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Standard XII

Mathematics

Sum of n Terms

The sum of th...

Question

The sum of the first n terms of the series, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + \dots . i s \frac{n}{2} (n + 1)^{2}$ , when n is even. When n is odd, the sum is

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Solution

The correct option is A
When n is odd, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + \dots . + 2 (n - 1)^{2} + n^{2}$
$= \frac{(n - 1) n^{2}}{2} + n^{2}$ , since (n - 1) is even. = $\frac{n^{2}}{2} (n + 1)$

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The sum of the first n terms of the series, 12+2.22+32+2.42+52+2.62+….isn2(n+1)2, when n is even. When n is odd, the sum is

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The sum of the first n terms of the series, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + 5^{2} + {2.6}^{2} + \dots . i s \frac{n}{2} (n + 1)^{2}$ , when n is even. When n is odd, the sum is

The correct option is A
When n is odd, $1^{2} + {2.2}^{2} + 3^{2} + {2.4}^{2} + \dots . + 2 (n - 1)^{2} + n^{2}$
$= \frac{(n - 1) n^{2}}{2} + n^{2}$ , since (n - 1) is even. = $\frac{n^{2}}{2} (n + 1)$