$1.3+3.5+5.7+....+(2n-1)(2n+1)=?

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Solution

$1.3 + 3.5 + 5.7 + (2 n - 1) (2 n + 1)$
$T_{n} = (2 n - 1) (2 n + 1) = 4 n^{2} - 1$
$S_{n} = \sum_{n} T_{n} = \sum_{n} (4 n^{2} - 1) = 4 \sum_{n} n^{2} - \sum_{n} 1 = \frac{4 n (n + 1) (2 n + 1)}{6} - n$
$S_{n} = \frac{2}{3} n (n + 1) (2 n + 1) - n = \frac{n (4 n^{2} + 6 n - 1)}{3}$

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Similar questions

1.3 + 3.5 + 5.7 + . . . + (2 n - 1) (2 n + 1) =

\frac{n (4 n^{2} + 6 n - 1)}{3}

(2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3}

$1.3 + 3.5 + 5.7 + \dots + (2 n - 1) (2 n + 1) = \frac{n (4 n^{2} + 6 n - 1)}{3} .$

1.3 + 3.5 + 5.7 + \dots (2 n - 1) (2 n + 1) = \frac{n (4 n + 6 n - 1)}{3}

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