Find the sum of the series ${1.3}^{2} + {2.5}^{2} + {3.7}^{2} +$ to $n$ terms.

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Solution

$\begin{matrix} S = {1.3}^{2} + {2.5}^{2} + {3.7}^{2} + . . . . . . S = \sum_{r = 1}^{n} r {(2 r + 1)}^{2} = \sum r (4 r^{2} + 1 + 4 r) = 4 \sum r^{3} + \sum r + 4 \sum r^{2} = \frac{4 {[n (n + 1)]}^{2}}{4} + \frac{n (n + 1)}{2} + 4 \frac{n (n + 1) (2 n + 1)}{6} = n^{2} {(n + 1)}^{2} + \frac{n (n + 1)}{2} + \frac{2}{3} (n) (n + 1) (2 n + 1) = n (n + 1) [n (n + 1) + \frac{1}{2} + \frac{2}{3} (2 n + 1)] = n (n + 1) [n^{2} + \frac{7 n}{3} + \frac{7}{6}] \end{matrix}$

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20

1. 3^{2} + 2. 5^{2} + 3. 7^{2} + . . .

Sum to 20 terms of the series 1.3² + 2.5² + 3.7² +... is:

{1.3}^{2} + {2.5}^{2} + {3.7}^{2} + . . . .

Find the sum to n terms of the series

1^{2} + 3^{2} + 5^{2} + \dots

n

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