The correct option is B 3n(n+1)(n+2) ∑ _ k=1 ^ n 3k(3k+3)=9∑ _ k=1 ^ n k^2 +9∑ _ k=1 ^ n k =9×n(n+1)(2n+1)6+9×n(n+1)2 =32n(n+1)(2n+1+3) ...

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Integration of Piecewise Continuous Functions

3.6+6.9+9.12+...

Question

$3.6 + 6.9 + 9.12 + . . . + 3 n (3 n + 3) =$

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

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3 n (n + 1) (n + 2)

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

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\frac{(n + 1) (n + 2) (n + 4)}{4}

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Solution

The correct option is B $3 n (n + 1) (n + 2)$
$\sum_{k = 1}^{n} 3 k (3 k + 3) = 9 \sum_{k = 1}^{n} k^{2} + 9 \sum_{k = 1}^{n} k$

$= 9 \times \frac{n (n + 1) (2 n + 1)}{6} + 9 \times \frac{n (n + 1)}{2}$

$= \frac{3}{2} n (n + 1) (2 n + 1 + 3)$

$= 3 n (n + 1) (n + 2)$

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3.6 + 6.9 + 9.12 + \dots + 3 n (3 n + 3) = 3 n (n + 1) (n + 2)

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