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

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

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\frac{n (3 n^{2} + 5 n + 1)}{3}

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\frac{n (5 n^{2} + 7 n - 1)}{3}

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\frac{n (7 n^{2} - 5 n + 1)}{3}

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Solution

The correct option is A $\frac{n (4 n^{2} + 6 n - 1)}{3}$
General term : $(2 k - 1) (2 k + 1)$

$\sum_{k = 1}^{n} (2 k - 1) (2 k + 1)$ $= \sum_{k = 1}^{n} (4 k^{2} - 1)$

$= \sum_{k = 1}^{n} (4 k^{2}) - \sum_{k = 1}^{n} (1)$

By using the formula,
$= 4 \frac{n (n + 1) (2 n + 1)}{6} - n$

On simplifying, we get the desired result !

Suggest Corrections

Similar questions

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

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

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

Prove that,

1.3+3.5+5.7+.............+(2n-1)(2n+1)=(n(4n²+6n-1))/3.

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

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