Question

# If $1+\frac{1+2}{2}+\frac{1+2+3}{3}+....$ to n terms is S, then S is equal to (a) $\frac{n\left(n+3\right)}{4}$ (b) $\frac{n\left(n+2\right)}{4}$ (c) $\frac{n\left(n+1\right)\left(n+2\right)}{6}$ (d) n2

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Solution

## $\left(\mathrm{a}\right)\frac{n\left(n+3\right)}{4}$ Let ${T}_{n}$ be the nth term of the given series. Thus, we have: ${T}_{n}=\frac{1+2+3+4+5+...+n}{n}=\frac{n\left(n+1\right)}{2n}=\frac{n}{2}+\frac{1}{2}$ Now, let ${S}_{n}$ be the sum of n terms of the given series. Thus, we have: ${S}_{n}=\sum _{k=1}^{n}\left(\frac{k}{2}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\sum _{k=1}^{n}\frac{k}{2}+\frac{n}{2}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+1\right)}{4}+\frac{n}{2}\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n}{2}\left(\frac{n+1}{2}+1\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n}{2}\left(\frac{n+3}{2}\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{n}=\frac{n\left(n+3\right)}{4}$

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