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

1 Answer

Oct 2, 2016

#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#.

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