Q43 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics

If $S_n$ is the sum of the first $n$ terms of the series $1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty,$ then, when $n$ is even $S_n=$

  • A
    $\frac{n(n+1)}{2}$
  • B
    $\frac{n^2(n+1)}{2}$
  • Correct answer
    C
    $\frac{n(n+1)^2}{2}$
  • D
    $\frac{n^2(n+2)}{2}$

Solution

We have, $S_n=1^2+2 \times 2^2+3^2+2 \times 4^2+5^2+2 \times 6^2+\ldots \infty$ $\therefore T_n=(2 n-1)^2+2 \times(2 n)^2$ $=4 n^2+1-4 n+8 n^2$ $=12 n^2-4 n+1$ $\Sigma T_n=12 \Sigma n^2-4 \Sigma n+\Sigma 1$ $=12 \frac{n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n$ $=2 n(n+1)(2 n+1)-2 n(n+1)+n$ $=2 n(n+1)(2 n+1-1)+n$ $=2 n(n+1)(2 n)+n$ $=4 n^2(n+1)+n=n[4 n(n+1)+1]$ $=n\left(4 n^2+4 n+1\right)=n(2 n+1)^2$ Here, $S_n$ is even $S_m=\frac{m}{2}\left(2 \frac{m}{2}+1\right)^2 n=\frac{m}{2} \Rightarrow S_m=\frac{m}{2}(m+1)^2$ $\therefore S_n=\frac{n}{2}(n+1)^2$ when $n$ is even

Hence, the correct answer is option: 'C'.

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