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I encountered the following problem in my textbook

Prove (1^3 + 2^3 + ... + n^3 ) = ( 1 + 2 +... + n) ^2.

Here is my solution, however I become stuck on the final part of the problem.

Let S1 = 1/2(n+1)

Let S2 = 1/6n(n+1)(2n+1)

S3 = (1^3 + 2^3 + ... + n^3)

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

= 4v^3 + 6v^2 + 4v + 1.

Substituting v = 0 , 1 , 2, ... , n into the equation and adding we get

4(0)^3 + 6(0)^2 + 4(0) + 14(1)^3 + 6(1)^2 + 4(1) + 14(2)^3 + 6(2)^2 + 4(2) +14(n)^3 + 6(n)^2 + 4(n) + 1

= 4S3 + 6S2 + 4S1 + n + 1

(n+1) ^4 = 4S3 + 6S2 + 4S1 + n + 1

Since we are solving for S3, we collect terms and get

4S3 = (n+1) ^4 - 6S2 - 4S1 - n - 1

= (n+1) ([n+1]^3 - 1- n(2n+1) - 2n)

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

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

= (n+1) (n^3 + n^2 )

S3 = ΒΌ (n+1)(n^3 + n^2)

Here is where I become stuck.

How does S3 = ( 1 + 2 +... + n) ^2?

Any help would be greatly appreciated.

Thanks

Also do you know where I can get LaTex?

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# Proof of a Sequence

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