SOLUTION: prove by mathematical induction: 1^5 + 2^5 + 3^5 + ... + n^5 = (1/12)n^2(n+1)^2(2n^2 + 2n -1)

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Question 1184007: prove by mathematical induction:
1^5 + 2^5 + 3^5 + ... + n^5 = (1/12)n^2(n+1)^2(2n^2 + 2n -1)

Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!
prove by mathematical induction:
1^5 + 2^5 + 3^5 + ... + n^5 = (1/12)n^2(n+1)^2(2n^2 + 2n -1)
--------

Base case: = 1
= = 1
Base case holds.

Hypothesis:
Assume + ... + = for n=k. (*)

Step case: Let n=k+1
[ + ... + ] +
... use hypothesis on terms within [ ] ...
= +
... this reduces to (you can factor then use a factoring website for the rest) ... (see NOTE)
=
The proof is complete here, but it is not obvious, so
let u=k+1 --> k=u-1
=
=
Proof complete
NOTE: I admit, there may be an easier path here. If other tutors (or the student) wish to find it, more power to them. I gave the general idea...
If this helps the student, thanks are appreciated.

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