(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Use mathematical induction to prove the formula for every positive integer n.

[tex]\sum_{i=1}^{5}i^5=\frac{n^2(n+1)^2(2n^2+2n-1)}{12}[/tex]

2. Relevant equations

3. The attempt at a solution

I know this will be true because the RHS is just your standard sums of powers of integers to the fifth formula, but nevertheless I have to prove it...

I know n=1 will work so I set n=k

[tex]\sum_{k=1}^{5}k^5=\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]

I take the RHS and set k=(k+1)

[tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2(k+1)-1)}{12}[/tex]

[tex]=\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]

If the test works for all the terms then it must be correct for the first 5 terms so I can eliminate the summation and just have the [tex]k^5[/tex]

so then I'd have

[tex](k+1)^5+\frac{k^2(k+1)^2(2k^2+2k-1)}{12}[/tex]

[tex]=\frac{k^2(k+1)^2(2k^2+2k-1)+12(k+1)^5}{12}[/tex]

From here I'm not sure how to get that to equal this:

[tex]\frac{(k+1)^2(k+2)^2(2(k+1)^2+2k+1)}{12}[/tex]

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# Homework Help: Mathematical Induction

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