How can I find this summation? I started by expanding $(k+1)^5$ and setting the summation of both equal to each other. There is some cancellation but I don't know what to do afterwards.
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do u know fourier series ? – wanderer Mar 18 '14 at 18:44
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No, I haven't learned any Fourier Series yet. – Jesus Mar 18 '14 at 18:46
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@Thomas Yes but I want to know how it is derived – Jesus Mar 18 '14 at 18:47
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You can see Faulhaber's Formula: http://en.wikipedia.org/wiki/Faulhaber's_formula (but this is overkill). – Batman Mar 18 '14 at 18:49
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@Batman batman always goes for overkill. no kill like overkill. – Guy Mar 18 '14 at 18:57
2 Answers
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HINT:
$$(k+1)^5-k^5=\binom51k^4+\binom52k^3+\binom53k^2+\binom54k+1$$
Set $k=1,2\cdots,n$ and add to find $$(n+1)^5-1=5S_4+10S_3+10S_2+5S_1+S_0$$
where $S_m=\sum_{r=1}^nr^m$
More generally, $$(k+1)^{n+1}-k^{n+1}=\sum_{r=0}^n\binom{n+1}rS_r$$
lab bhattacharjee
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The summation of both sides, in equation, would also be to true right? Then I would just need to solve for $S_4$? How would I do that? – Jesus Mar 18 '14 at 19:01
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@Jesus $S_3$ is the sum of cubes. $S_2$ is the sum of squares. $S_1$ is the sum of natural numbers. You already know those formulas, right? – Guy Mar 18 '14 at 19:02
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Yes, so I would just need to substract all of the terms to leave $5S_4$, then divide by 5? – Jesus Mar 18 '14 at 19:05
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@Jesus, as Sabyasachi has suggested, we need to know the sums of all the lower powers – lab bhattacharjee Mar 19 '14 at 03:12
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One way that should be easy to remember is just linear algebra. it is guaranteed that the answer is $$ a_5 n^5 + a_4 n^4 + a_3 n^3 + a_2 n^2 + a_1 n + a_0, $$ where all six $a_j$ are rational numbers. They might all be positive, I'm not sure, but $a_5 > 0.$
So, the sum up to $n=1$ is $1,$ or $$ a_5 + a_4 + a_3 + a_2 + a_1 + a_0 = 1. $$
The sum up to $n=2$ is $17,$ or $$ 32 a_5 + 16 a_4 + 8 a_3 + 4 a_2 + 2 a_1 + a_0 = 17. $$
The sum up to $n=3$ is $98,$ or $$ 243 a_5 + 81 a_4 + 27 a_3 + 9 a_2 + 3 a_1 + a_0 = 98. $$
The sum up to $n=4$ is $354,$ or $$ 1024 a_5 + 256 a_4 + 64 a_3 + 16 a_2 + 4 a_1 + a_0 = 354. $$
The sum up to $n=5$ is $979,$ or $$ 3125 a_5 + 625 a_4 + 125 a_3 + 25 a_2 + 5 a_1 + a_0 = 979. $$
The sum up to $n=6$ is $2275,$ or $$ 7776 a_5 + 1296 a_4 + 216 a_3 + 36 a_2 + 6 a_1 + a_0 = 2275. $$
So, there you go, six linear equations in six unknowns. You just solve the system with Gauss-Jordan elimination. I saw one of your questions was on row echelon form, so you know how to do this.
There are also many clever ways to do this. Matter of taste.
Will Jagy
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Don't you think that $a_0$ is obvious ? If I am correct, one less equation. In fact, I wonder how easily could we prove that for the sum of $k^i$,$a_{i+1}=1/(i+1)$ and $a_i=1/2$. Cheers. – Claude Leibovici Mar 20 '14 at 07:55
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@ClaudeLeibovici, yes, $a_{i+1} = 1 / (i+1)$ follows from methods of finite differences. Not sure about the next one. i agree that $a_0 = 0$ because the sum of $0$ terms is $0,$ but decided to leave that out as needing explanation. I imagine a good deal can be shown just by induction on the exponent. – Will Jagy Mar 20 '14 at 18:09