- #1
themadhatter1
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Homework Statement
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]
Homework Equations
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]