# How do you find the sum of the series 4n^3 from n=1 to n=3?

Oct 2, 2016

$144.$

#### Explanation:

We will use the following Formula :

$: {\sum}_{j = 1}^{j = n} {j}^{3} = \sum {n}^{3} = {1}^{3} + {2}^{3} + {3}^{3} + \ldots \ldots + {n}^{3} = \frac{{n}^{2} {\left(n + 1\right)}^{2}}{4.}$

Reqd. Sum$= {\sum}_{n = 1}^{n = 3} 4 {n}^{3} = 4 {\sum}_{n = 1}^{n = 3} {n}^{3}$

$= 4 \left\{\frac{\left({3}^{2}\right) {\left(3 + 1\right)}^{2}}{4}\right\} = \left(9\right) \left(16\right) = 144$.