# proof of $1^4+2^4+...+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ [duplicate]

I want a 'simple' proof to show that:

$$1^4+2^4+...+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$

I tried to prove it like the others but I can't and now I really need the proof. Also I want a geometric proof for that and this one below:

$$1^3+2^3+3^3+...=(\frac{n(n+1)}{2})^2$$

geometric proof: I want a proof using shapes and geometry.

• Both can easily be proved by induction. Jun 8, 2016 at 11:53
• What you mean by geometric proof ? Jun 8, 2016 at 11:57
• For the second one: math.stackexchange.com/questions/61482/… Jun 8, 2016 at 12:05
• Already answered (in general) at math.stackexchange.com/questions/1801457/… Jun 8, 2016 at 12:16
• I mean that using shapes and geometry. Jun 8, 2016 at 14:29

$$\sum_{i=1}^n i^3=\left(\sum_{i=1}^n i\right)^2$$
\begin{align} & \sum\limits_{i=1}^{n}{{{(i+1)}^{5}}-{{i}^{5}}}={{(n+1)}^{5}}-{{1}^{5}} \\ & \sum\limits_{i=1}^{n}{5{{i}^{4}}+10{{i}^{3}}+10{{i}^{2}}+5i+1=}{{(n+1)}^{5}}-{{1}^{5}} \\ & 5\sum\limits_{i=1}^{n}{{{i}^{4}}+10\sum\limits_{i=1}^{n}{{{i}^{3}}+10\sum\limits_{i=1}^{n}{{{i}^{2}}+5\sum\limits_{i=1}^{n}{i\,\,+\,n=}}}}{{(n+1)}^{5}}-{{1}^{5}} \\ & 5{{\sum\limits_{i=1}^{n}{{{i}^{4}}+10\left( \frac{n(n+1)}{2} \right)}}^{2}}+10\left( \frac{n(n+1)(2n+1)}{6} \right)+5\left( \frac{n(n+1)}{2} \right)+n={{(n+1)}^{5}}-{{1}^{5}} \\ \end{align} by simplification,we have $$\sum\limits_{i=1}^{n}{{{i}^{4}}}=\frac{n(n+1)(2n+1)(3{{n}^{2}}+3n-1)}{30}$$
If you want "a simple proof" you can calculate the difference $$(n+1)(n+2)(2n+3)(3n^2+9n+5)-n(n+1)(2n+1)(3n^2+3n-1)$$ where the greater quantity comes from putting $n+1$ in the given formula. You can verify this is equal to $30(n+1)^4$ (you need for this some school calculation and you have this way determined the expression of $n(n+1)(2n+1)(3n^2+3n-1)+(n+1)^4$ ). And you can finish applying an immediate induction.