Find the sum to n terms of the series whose n | vedclass

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(B) Given that the $n^{	ext{th}}$ term is $a_n = n(n+3) = n^2 + 3n$.The sum to $n$ terms is given by $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + 3k

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

Vedclass · Answer

(B) Given that the $n^{	ext{th}}$ term is $a_n = n(n+3) = n^2 + 3n$.The sum to $n$ terms is given by $S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n (k^2 + 3k)$.Using the standard summation formulas $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,we get:$S_n = \frac{n(n+1)(2n+1)}{6} + 3 	imes \frac{n(n+1)}{2}$.Taking $\frac{n(n+1)}{2}$ as a common factor:$S_n = \frac{n(n+1)}{2} \left[ \frac{2n+1}{3} + 3 ight] = \frac{n(n+1)}{2} \left[ \frac{2n+1+9}{3} ight] = \frac{n(n+1)(2n+10)}{6}$.$S_n = \frac{n(n+1) 	imes 2(n+5)}{6} = \frac{n(n+1)(n+5)}{3}$.

Find the sum to $n$ terms of the series whose $n^{\text{th}}$ term is $n(n+3)$.

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