The sum of first n terms of the series 12+2×22+32+2×42+52+2×62+… is n(n+1)2/2when n is even. When n is odd the sum of the series is

Let n=2m, thenS2m=12+2×22+32+2×42+…+(2m−1)2+2(2m)2=2m(2m+1)2/2=m(2m+1)2When n = 2m – 1,S2m−1=12+2×22+32+2×42+…+=S2m−2(2m)2=m(2m+1)2−2(2m)2=m4m2+4m+1−8m=m(2m−1)2=n2(n+1)/2

The sum of first n terms of the series 12+2×22+32+2×42+52+2×62+… is n(n+1)2/2when n is even. When n is odd the sum of the series is

Let n=2m, thenS2m=12+2×22+32+2×42+…+(2m−1)2+2(2m)2=2m(2m+1)2/2=m(2m+1)2When n = 2m – 1,S2m−1=12+2×22+32+2×42+…+=S2m−2(2m)2=m(2m+1)2−2(2m)2=m4m2+4m+1−8m=m(2m−1)2=n2(n+1)/2

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

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$n^{2} (n + 1) / 2$

none of these

$n^{2} (3 n + 1) / 4$

$n^{3} (n - 1) / 2$

answer is B.

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Detailed Solution

Let $n = 2 m,$ then

$\begin{aligned} S_{2 m} = 1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + \dots + \\ (2 m - 1)^{2} + 2 (2 m)^{2} \\ = 2 m (2 m + 1)^{2} / 2 = m (2 m + 1)^{2} \end{aligned}$

When $n = 2 m - 1,$

$\begin{array}{l} S_{2 m - 1} = 1^{2} + 2 \times 2^{2} + 3^{2} + 2 \times 4^{2} + \dots + \\ = S_{2 m} - 2 (2 m)^{2} = m (2 m + 1)^{2} - 2 (2 m)^{2} \\ = m ⌊4 m^{2} + 4 m + 1 - 8 m⌋ = m (2 m - 1)^{2} \\ = n^{2} (n + 1) / 2 \end{array}$