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By considering $$\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)$$ Show that $$\sum _{r=1}^n r^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2 + 3n-1)$$

So I'm a little unsure on how to really start on this so I've expanded the given expression an found that $$\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right) = \sum_{r=1} ^n 5r^4 +10r^3 + 10r^2 +5r+1$$

I know I want to get this into a form which I can then solve for $\sum _{r=1}^n r^4 $ using the fact I know the sums of $\sum _{r=1}^n r^3$ and $\sum _{r=1}^n r^2$. If anyone could helping me get started with this question that would be great.

H.Linkhorn
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5 Answers5

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Define $S_k:=\sum_{r=1}^n n^k$ so you've shown $(n+1)^5-1=5S_4+10S_3+10S_2+5S_1+S_0$. This allows us to express $S_4$ in terms of the $S_i$ with $0\le i<4$. You could of course have used a similar approach to derives relations between earlier $S_i$. From these we can gradually prove$$S_0=n,\,S_1=\frac12n(n+1),\,S_2=\frac{1}{6}n(n+1)(2n+1),\,S_3=\frac{1}{4}n^2(n+1)^2.$$Hence $$5S_4=n\left(n^4+5n^3+10n^2+10n+5-\frac{5}{2}n(n+1)^2-\frac{5}{3}(n+1)(2n+1)-\frac{5}{2}(n+1)-1\right).$$After that it's just tedious algebra.

J.G.
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HINT:

  1. $$\small\sum _{r=1} ^n \left( (r+1)^5 - r^5 \right)=(2^5-1^5)+(3^5-2^5)+\cdots+(n^5+(n-1)^5)+((n+1)^5-n^5)=\,?$$

  2. $$\sum_{r=1}^n r^3=\left(\sum_{r=1}^n r\right)^2,\quad \sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}6,\quad \sum_{r=1}^n r=\frac{n(n+1)}2$$

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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OK, i will also type the "same answer", but am not that quick...

We use below a natural $n$. Let $S(n,p)$ be the sum of the numbers $1^p, 2^p,\dots, n^p$. Then: $$ \begin{aligned} (n+1)^5-1^5 &=\sum_{1\le r\le n}((r+1)^5-r^5)\\ &=5S(n,4)+10S(n,3)3+10S(n,2)+5S(n,1)+S(n,0)\\ &=5S(n,4) +\frac {10}4\cdot n^2(n+1)^2 +\frac {10}6\cdot n(n+1)(2n+1) +\frac 52\cdot n(n+1) +n \\[2mm] &\qquad\text{ so} \\[2mm] 30S(n,4) &= 6((n+1)^5-1) -15 n^2(n+1)^2 -10 n(n+1)(2n+1) -15 n(n+1) -6 n \\ &= 6n(n^4+5n^3+10n^2+10n+5) -15 n^2(n+1)^2 -10 n(n+1)(2n+1) -15 n(n+1) -6 n \\ &= 6n(n^4+5n^3+10n^2+10n+4) -15 n^2(n+1)^2 -10 n(n+1)(2n+1) -15 n(n+1) \\ &= 6n(n+1)(n^3+4n^2+6n+4) -15 n^2(n+1)^2 -10 n(n+1)(2n+1) -15 n(n+1) \\ &= n(n+1)\Big(\ 6(n^3+4n^2+6n+4)-15(n^2+n)-10(2n+1)-15\ \Big) \\ &= n(n+1)\Big(\ 3(n^2 + n + 3)(2n + 1)-10(2n+1)\ \Big) \\ &= n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 9-10\ \Big) \\ &= n(n+1)(2n+1)\Big(\ 3n^2 + 3n + 1\ \Big) \end{aligned} $$

dan_fulea
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  • Where does the very first line come from $$(n+1)^5 -1 =\sum ((r+1)^5 -1)$$ – H.Linkhorn Feb 23 '19 at 08:15
  • We have a telescopic sum, for $n=7$ to get explicitly the idea in case of an example, we have the sum (in my prefered order) $$(8^5-7^5)+(7^5-6^5)+(6^5-5^5)+(5^5-4^5)+(4^5-3^5)+(3^5-2^5)+(2^5-1^5)\ ,$$ and only the terms $(n+1)^5=8^5$ in our case and $1^5$ (with a minus sign) survive in the final after the telescopic cancellations. – dan_fulea Feb 25 '19 at 12:57
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You're on the right track. Write these relations for $r=n, n-1,\dots ,2,1$ and add them. You'll have a telescoping sum on the l.h.s. and ultimately, denoting $S_k=\sum_{r=1}^n r^k $, you'll obtain a recursive relation between $S_4, S_3, S_2$ and $S_1$. The last three are supposed to be known, so you'll be able to deduce $S_4$.

Bernard
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You can split up sums: $$\sum(a+b)=\sum a+\sum b$$ So use that:$$\sum(r^4)=\frac 15\bigg[\sum((r+1)^5-r^5)-10\sum(r^3)-10\sum(r^2)-5\sum r-1\bigg]$$

Rhys Hughes
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