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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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nth n^{\text {th }} term of the series 122+12+224+12+22+326+……\frac{1^{2}}{2}+\frac{1^{2}+2^{2}}{4}+\frac{1^{2}+2^{2}+3^{2}}{6}+\ldots \ldots is
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Correct option: C
(n+1)(2n+1)12\frac{(n+1)(2 n+1)}{12}
Option C ((n+1)(2n+1)/12) is correct because it represents the nth term of the series, derived by summing the squares of the first n natural numbers and dividing by 2n.

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