to n terms is S, then S is equal to


If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+\ldots$ to $n$ terms is $S$, then $S$ is equal to

(a) $\frac{n(n+3)}{4}$

(b) $\frac{n(n+2)}{4}$

(c) $\frac{n(n+1)(n+2)}{6}$

(d) $n^{2}$


(a) $\frac{n(n+3)}{4}$

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=\frac{1+2+3+4+5+\ldots+n}{n}=\frac{n(n+1)}{2 n}=\frac{n}{2}+\frac{1}{2}$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:


$\Rightarrow S_{n}=\sum_{k=1}^{n} \frac{k}{2}+\frac{n}{2}$

$\Rightarrow S_{n}=\frac{n(n+1)}{4}+\frac{n}{2}$

$\Rightarrow S_{n}=\frac{n}{2}\left(\frac{n+1}{2}+1\right)$

$\Rightarrow S_{n}=\frac{n}{2}\left(\frac{n+3}{2}\right)$

$\Rightarrow S_{n}=\frac{n(n+3)}{4}$



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