The correct option is B n(n+1)(n+5)/3 ∑ _ k=1 ^ n k(k+3)=∑ _ k=1 ^ n k^2 +3∑ _ k=1 ^ n k = n(n+1)(2n+1)6+3×n(n+1)2 =n(n+1)2(2n+13+3) = ...

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Integration of Piecewise Continuous Functions

1.4+2.5+...+n...

Question

$1.4 + 2.5 + . . . + n (n + 3) =$

\frac{n (n + 3) (n + 5)}{9}

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

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\frac{n (n + 5) (n + 7)}{6}

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\frac{n (n + 3) (n + 9)}{12}

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Solution

The correct option is B $\frac{n (n + 1) (n + 5)}{3}$
$\sum_{k = 1}^{n} k (k + 3) = \sum_{k = 1}^{n} k^{2} + 3 \sum_{k = 1}^{n} k$
$= \frac{n (n + 1) (2 n + 1)}{6} + 3 \times \frac{n (n + 1)}{2}$
$= \frac{n (n + 1)}{2} (\frac{2 n + 1}{3} + 3)$
$= \frac{n (n + 1) (n + 5)}{3}$

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n - \frac{n (n - 1) (n - 2)}{⌊ 3} \cdot \frac{1}{3} + \frac{n (n - 1) (n - 2) (n - 3) (n - 4)}{⌊ 5} \cdot \frac{1}{3^{2}} - . . . . .,

lim n \to \infty \frac{1^{1} + 2^{2} + 3^{3} + \dots + n^{n}}{1^{n} + 2^{n} + 3^{n} + \dots + n^{n}}

1.2.3 + 2.3.4 + … + n(n + 1) (n + 2) =

(1 \times 2 \times 5) + (2 \times 3 \times 7) + (3 \times 4 \times 9) + . . . . .

\frac{n (n + 1) (n + 2) (3 n + 7)}{6}

1.2.3 + 2.3.4 + \dots + n (n + 1) (n + 2) = \frac{n (n + 1) (n + 2) (n + 3)}{12}

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