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+... is (n+1)^22 when n is even. When n is odd the sum is - Mathematics MCQ AIEEE 2004 2004 | ExamTest.live

hardMCQAIEEE 20042004Mathematics

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The sum of the first n terms of the series 1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + ... is \frac{n ( n + 1 ) ^{2}}{2} when n is even. When n is odd the sum is

Solution & Step-by-step Explanation

Let S_{n} be the sum of first n terms. Given for even n, S_{n} = \frac{n ( n + 1 ) ^{2}}{2} . When n is odd, (n - 1) is even. The sum of first n terms is S_{n} = S_{n - 1} + T_{n} . Since n is odd, the n th term T_{n} = n^{2} . S_{n} = \frac{( n - 1 ) (( n - 1 ) + 1 ) ^{2}}{2} + n^{2} . S_{n} = \frac{( n - 1 ) n ^{2}}{2} + n^{2} . S_{n} = n^{2} [\frac{n - 1}{2} + 1] S_{n} = n^{2} [\frac{n - 1 + 2}{2}] = \frac{n ^{2} ( n + 1 )}{2} .

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The sum of the first n terms of the series 1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + ... is \frac{n ( n + 1 ) ^{2}}{2} when n is even. When n is odd the sum is

\frac{3 n ( n + 1 )}{2}

\frac{n ^{2} ( n + 1 )}{2}

\frac{n ( n + 1 ) ^{2}}{4}

[\frac{n ( n + 1 )}{2}]^{2}

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