# If nth term of a series is 1/2 ×n(n+1) then find the sum of n terms of the series ?

Aug 7, 2018

${S}_{n} = \frac{n}{6} \left(n + 1\right) \left(n + 2\right)$

#### Explanation:

Here,

${n}^{t h} t e r m$ of series is :$\frac{n}{2} \left(n + 1\right)$

So, the sum of first terms of series is:

${S}_{n} = 1 + \left(1 + 2\right) + \left(1 + 2 + 3\right) + \ldots + \frac{n}{2} \left(n + 1\right)$

$\therefore {S}_{n} = {\sum}_{r = 1}^{n} \frac{r}{2} \left(r + 1\right) = \frac{1}{2} {\sum}_{r = 1}^{n} \left({r}^{2} + r\right)$

$\therefore {S}_{n} = \frac{1}{2} \left\{{\sum}_{r = 1}^{n} {r}^{2} + {\sum}_{r = 1}^{n} r\right\}$

$\therefore {S}_{n} = \frac{1}{2} \left\{\frac{n}{6} \left(n + 1\right) \left(2 n + 1\right) + \frac{n}{2} \left(n + 1\right)\right\}$

$\therefore {S}_{n} = \frac{1}{2} \cdot \frac{n}{2} \left(n + 1\right) \left\{\frac{2 n + 1}{3} + 1\right\}$

$\therefore {S}_{n} = \frac{n}{4} \left(n + 1\right) \left\{\frac{2 n + 1 + 3}{3}\right\}$

$\therefore {S}_{n} = \frac{n}{12} \left(n + 1\right) \left\{2 n + 4\right\}$

$\therefore {S}_{n} = \frac{n}{12} \left(n + 1\right) 2 \left(n + 2\right)$

$\therefore {S}_{n} = \frac{n}{6} \left(n + 1\right) \left(n + 2\right)$