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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Application +4 | -1
If the sum of the series 12+2.22+32+2.42+52+…2.62+…1^{2}+2.2^{2}+3^{2}+2.4^{2}+5^{2}+\ldots 2.6^{2}+\ldots upto n\mathrm{n} terms, when n\mathrm{n} is even, is n(n+1)22\frac{n(n+1)^{2}}{2}, then the sum of the series, when n\mathrm{n} is odd, is
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Correct option: C
n2(n+1)2\frac{n^{2}(n+1)}{2}
The correct answer is 3 because the series can be simplified by rationalizing each term, leading to a telescoping series where most terms cancel out, leaving only the first and last terms.

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