Question

Prove the sum of cubes sequence

1^3+2^3+3^3+...+n^3 = \Big[ \cfrac{n(n+1)}{2}\Big] ^2

Collected in the board: Famous Math Sequences

Steven Zheng posted 1 day ago

Answer

Let

S_n = 1^3+2^3+3^3+...+n^3

First, let's evaluate the difference of a natural number to the 4th power

(n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]

=(2n^2+2n+1)(2n+1)

=4n^3+6n^2+4n+1

2^4-1^4=4×1^3+6×1^2+4*1+1

3^4-2^4=4×2^3+6×2^2+4*2+1

4^4-3^4=4×3^3+6×3^2+4*3+1

\dots

(n+1)^4-n^4=4n^3+6n^2+4n+1

Sum of all equations above gives

(n+1)^4 - 1 = 4S_n+6×(1^2+2^2+\dots+n^2) +4×(1+2+\dots+n) +n

Then solve for S_n

S_n =\cfrac{1}{4} [(n+1)^4 - 1 - 6×\cfrac{n(n+1)(2n+1)}{6} -2n(n+1) -n]

=\cfrac{n+1}{4}[(n+1)^3 -n(2n+1) - 2n-1]

=\cfrac{n+1}{4}[(n+1)^3 -(2n^2+3n+1)]

=\cfrac{n+1}{4}[(n+1)^3 -(2n+1)(n+1)]

=\cfrac{(n+1)^2}{4}[(n+1)^2 -(2n+1)]

=\cfrac{n^2(n+1)^2}{4} =\Big[ \cfrac{n(n+1)}{2}\Big] ^2

Steven Zheng posted 1 day ago

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