The sum 1 2 - 2 · 3 2 + 3 · 5 2 - 4 · 7 2 + 5 · 9 2 - . . . . . + 15 · 29 2 is _______. [2023]

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Solution

Q.
The sum $1^{2} - 2 \cdot 3^{2} + 3 \cdot 5^{2} - 4 \cdot 7^{2} + 5 \cdot 9^{2} - . . . . . + 15 \cdot 29^{2}$ is _______. [2023]

Ans.
(6952)

$Separating odd and even placed terms:$

$S = (1 \cdot 1^{2} + 3 \cdot 5^{2} + \dots + 15 \cdot {(29)}^{2}) - (2 \cdot 3^{2} + 4 \cdot 7^{2} + \dots + 14 \cdot {(27)}^{2})$

$S = \sum_{n = 1}^{8} (2 n - 1) {(4 n - 3)}^{2} - \sum_{n = 1}^{7} (2 n) {(4 n - 1)}^{2}$

$S = \sum_{n = 1}^{8} (32 n^{3} - 64 n^{2} + 42 n - 9) - \sum_{n = 1}^{7} (32 n^{3} - 16 n^{2} + 2 n)$

$S = 32 \sum_{n = 1}^{8} n^{3} - 64 \sum_{n = 1}^{8} n^{2} + 42 \sum_{n = 1}^{8} n - 9 \sum_{n = 1}^{8} (1) - 32 \sum_{n = 1}^{7} n^{3} + 16 \sum_{n = 1}^{7} n^{2} - 2 \sum_{n = 1}^{7} n$

$S = 32 {[\frac{n (n + 1)}{2}]}^{2} - 64 [\frac{n (n + 1) (2 n + 1)}{6}] + 42 [\frac{n (n + 1)}{2}] - 9 \times 8$ $- 32 {[\frac{n (n + 1)}{2}]}^{2} + 16 [\frac{n (n + 1) (2 n + 1)}{6}] - 2 [\frac{n (n + 1)}{2}]$

$S = 32 {[36]}^{2} - 64 [12 \times 17] + 42 [36] - 72 - 32 {(28)}^{2} + 16 (140) - 2 (28)$

$S = 6952$

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