The sum to first ' $n$ ' terms of the series $(n \geq 3)1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2

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The sum to first ' $n$ ' terms of the series $(n \geq 3)1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + \ldots +$ upto ' $n$ ' terms is equal to

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Sol. [Key Concept: Series sum splits into odd/even cases.]If n is even$$\begin{matrix} S & \ = 1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + \ldots + 2 \cdot n^{2} \ & \ = \sum_{r = 1}^{n}\mspace{2mu}\mspace{2mu} r^{2} + \sum_{r = 1}^{n/2}\mspace{2mu}\mspace{2mu}(2r)^{2} = \frac{n(n + 1)(2n + 1)}{6} + 4\frac{\frac{n}{2}\left( \frac{n}{2} + 1 ight)(n + 1)}{6} \ & \ = \frac{n(n + 1)^{2}}{2} \end{matrix}$$If n is odd$$S = \frac{(n - 1)n^{2}}{2} + n^{2} = \frac{n^{2}(n + 1)}{2}$$✔ (a) $n$ odd ⇒ $S = \frac{n^{2(}n + 1)}{2}$✔ (b) $n$ even ⇒ $S = \frac{n(n{+ 1)}^{2}}{2}$✘ (c) wrong parity/value✘ (d) wrong parity/value

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