Question

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

Answer 1

`144.`

Explanation:

We will use the following Formula :

`: sum_(j=1)^(j=n) j^3=sumn^3=1^3+2^3+3^3+... ...+n^3=(n^2(n+1)^2)/4.`

Reqd. Sum`=sum_(n=1)^(n=3) 4n^3=4sum_(n=1)^(n=3) n^3`

`=4{((3^2)(3+1)^2)/4}=(9)(16)=144`.