1

How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?

I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.

Skm
  • 2,296
  • By induction?? What other option is there? And if the algebra is complex, look for ways to simplify. – Mark Bennet Nov 11 '17 at 20:52
  • I'm sure that there are other ways to solve it. But in this particular instance, I'd like to solve by induction. As for the second statement, that's honestly easier said then done. I've spent a long time on this problem, so I think that's it more efficient to search for an answer. – Skm Nov 11 '17 at 20:56
  • Hah that was the exact link I gave :) …… wait, I did not do that comment! The one just above this I mean... who did that?? ;\ – Mr Pie Jan 16 '18 at 09:51

4 Answers4

1

your formula can be simplified to $$1/30\,n \left( 2\,n+1 \right) \left( n+1 \right) \left( 3\,{n}^{2}+3 \,n-1 \right) $$ for $n=1$ we have $$1^4=\frac{1}{30}\cdot 3\cdot 2\cdot 5$$ this is right now we assume that $$1^4+2^4+3^4+...+k^4=\frac{1}{30}k(2k+1)(k+1)(3k^2+3k-1)$$ now we have to prove that $$1+2^4+3^4+...+k^4+(k+1)^4=\frac{1}{30}(k+1)(2k+3)(k+2)(3(k+1)^2+3(k+1)-1)$$ have you got it? now you have to Show that $$(k+1)^4+\frac{1}{30}k(2k+1)(k+1)(3k^2+3k-1)=\frac{1}{30}(k+1)(2k+3)(k+2)(3(k+1)^2+3(k+1)-1)$$ can you finish?

Dr. Sonnhard Graubner
  • 95,283
1

The usual trick is to evaluate the sum $$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$ using $$k(k+1)(k+2)(k+3) = \dfrac{1}{5}k(k+1)(k+2)(k+3)(k+4) - \dfrac{1}{5}(k-1)k(k+1)(k+2)(k+3).$$

Then simply expand the summands to obtain $\sum_{k=1}^n k^4$.

dezdichado
  • 13,888
1

We have \begin{eqnarray*} 1^4+\cdots + n^4 &=& \frac{1}{5} n^5+\frac{1}{2}n^4+\frac{1}{3}n^3+\frac{1}{30}n) \\ &=& \frac{1}{30}(6 n^5+15n^4+10n^3-n). \\ \end{eqnarray*} So \begin{eqnarray*} 1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 n^5+15n^4+10n^3-n +30n^4+120n^3+180n^2+120n+30). \\ \end{eqnarray*} Now check this really does equal the expression below \begin{eqnarray*} \frac{1}{30}(6 (n^5 +5n^4+10n^3+10n^2+5n+1) \\ +15(n^4+4n^3+6n^2+4n+1) \\ +10(n^3+3n^2+3n+1) \\ -(n+1) ). \\ \end{eqnarray*} So \begin{eqnarray*} 1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 (n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)). \\ \end{eqnarray*}

Donald Splutterwit
  • 36,613
  • 2
  • 26
  • 73
1

$$\underbrace{1^4+2^4+3^4\cdots\ + n^4}_{\sum_{k = 1}^n k^4} = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$$ See the Faulhaber's formula and the Bernoulli number: $\text{B}_{j}(x)$

\begin{align} \sum_{k=0}^{n} k^p = \frac{1}{p+1} \sum_{j=0}^p (-1)^j {p+1 \choose j} B_j n^{p+1-j}=\frac{\text{B}_{p+1}(n+1)-\text{B}_{p+1}(0)}{p+1} \end{align}

In your case \begin{align} \sum_{k = 1}^n k^4 &= \frac{\text{B}_{4+1}(n+1)-\text{B}_{4+1}(0)}{4+1}\\ &= \frac{1}{5}\text{B}_{5}(n+1)\\ &=\frac{1}{5}\left(\frac16 (-1 - n) + \frac53 (1 + n)^3 - \frac52 (1 + n)^4 + (1 + n)^5\right)\\ &=\frac{1}{5}\left(\frac{\left(-1-n\right)}{6}+\frac{5\left(1+n\right)^3\cdot 2}{6}-\frac{5\left(1+n\right)^4\cdot 3}{6}+\frac{\left(1+n\right)^5\cdot 6}{6}\right)\\ &=\frac{1}{5}\left(\frac{6n^5+15n^4+10n^3-n}{6}\right)\\ &=\frac{6n^5+15n^4+10n^3-n}{30}\\ &= \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n \tag*{$\Box$} \end{align}

Mr Pie
  • 9,459
Darío A. Gutiérrez
  • 3,863
  • Hah I though u tagged the box $\Box$ without the brackets and so I thought to myself, “How did you do that?” Then when I click on edit to see what you wrote, I just see \quad\quad\quad\quad\quad... haha you can also use \qquad if you like :) – Mr Pie Jan 16 '18 at 06:03
  • \qquad it does not work :( . could you edit it ? – Darío A. Gutiérrez Jan 16 '18 at 07:22
  • I will edit it. After asking a question relating to the matter on the meta.stackexchange, I will edit it :) – Mr Pie Jan 16 '18 at 09:49
  • Thanks you!!!!! – Darío A. Gutiérrez Jan 16 '18 at 09:53
  • Wow that was quick!! Sorry for the delay; I had to go out somewhere, so I decided to just post on the math.meta and when I get back, I’ll have the full details. No worries though ;) … Oh and in case you don’t know, I just did \tag*{$$$\Box$$$} – Mr Pie Jan 16 '18 at 09:54