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

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Answer 1 · 2016-10-02T10:51:28

$144.$

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 $=sum_(n=1)^(n=3) 4n^3=4sum_(n=1)^(n=3) n^3$

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

How do you find the sum of the series 4n^34n^3 from n=1 to n=3?