Consider the following:




In General is it true for further increase i.e.,


$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$

  • 1
    $\begingroup$ Could you prove this by induction? $\endgroup$ Dec 15 '17 at 17:42
  • 2
    $\begingroup$ The way to proof a statement like this is mathematical induction, that is: you have checked the expresion holds for lower numbers. Now suppose it is true also for a big number $n$ and try to show that the formula holds for $n+1$. The computotions are quite cumbersome. I'll try to give you an answer in a few minutes. But you can also try it. $\endgroup$
    – Dog_69
    Dec 15 '17 at 17:44

Both sides are polynomials in $n$ of degree $8$. Since they coincide for $n=0,\dots,8$, they are equal.

Any $9$ points will do. Taking $n=-4,\dots,4$ is probably easier to do by hand.

  • $\begingroup$ You are correct. My mistake! $\endgroup$ Dec 15 '17 at 20:10
  • $\begingroup$ how do you know beforehand that the polynomial $\sum_{i=1}^n i^5+i^7$ is of degree $8?$ The RHS is of degree 8 I agree. But assume we don't have any clue about the equality. $\endgroup$
    – Guy Fsone
    Dec 18 '17 at 14:05
  • $\begingroup$ @GuyFsone, the repeated differences of a polynomial of degree $7$ are zero after the $8$-th difference. So the repeated differences of the partial sums of a polynomial of degree $7$are zero after the $9$-th difference. $\endgroup$
    – lhf
    Dec 18 '17 at 15:38

The formula is already true for $n=1,2,....,5$ and we know that $$ \sum_{i=1}^ni= \frac{n\left(n+1\right)}{2}$$ Assume $$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4 =\frac{n^4\left(n+1\right)^4}{8}$$

then, $$\begin{align}\sum_{i=1}^{n+1} i^5+i^7&=\sum_{i=1}^{n} i^5+i^7 +(n+1)^5 +(n+1)^7\\&=\color{blue}{\frac{n^4\left(n+1\right)^4}{8}}+(n+1)^5 +(n+1)^7 \\&=\color{blue}{\frac{n^4\left(n+1\right)^4}{8}}+(n+1)^4\left[n+1 +(n+1)^3 \right] \\&=(n+1)^4\left( \frac{n^4}{8} +n+1 +\color{red}{n^3+3n^2+3n+1}\right) \\&=(n+1)^4\left( \frac{n^4}{8}+ n^3+3n^2+4n+2\right) \\&=\frac{(n+1)^4}{8}\left( n^4+ \color{blue}{4}\cdot\color{red}{2}\cdot n^3+\color{blue}{6}\cdot\color{red}{2^2}\cdot n^2+\color{blue}{4}\cdot\color{red}{2^3}\cdot n+\color{red}{2^4}\right) \\&=\color{blue}{\frac{\left(n+1\right)^4\left(n+2\right)^4}{8}=2\left( \sum_{i=1}^{n+1}i\right)^4}\end{align}$$

which prove that the formula is true

  • $\begingroup$ Ohh, nice trick. I have expanded all terms and then I have compared with $(n+1)^4(n+2)^4/8$. Yor way is much better. $\endgroup$
    – Dog_69
    Dec 15 '17 at 18:16
  • $\begingroup$ In the brackets, you should have $n + 1 + (n + 1)^{\color{red}{3}}$. $\endgroup$ Dec 16 '17 at 11:02
  • $\begingroup$ I like how you demonstrated the binomial expansion in your proof. It is very clearly presented. $\endgroup$ Dec 16 '17 at 15:50

Notice $\sum_{k=1}^n k = \frac{n(n+1)}{2}$. For the identity at hand,

$$\sum_{k=1}^n k^5 + \sum_{k=1}^n k^7 \stackrel{?}{=} 2 \left(\sum_{k=1}^n k\right)^4$$ If one compute the difference of successive terms in RHS, we find

$$\begin{align}{\rm RHS}_n - {\rm RHS}_{n-1} &= 2\left(\frac{n(n+1)}{2}\right)^4-2\left(\frac{n(n-1)}{2}\right)^4\\ &= \frac{n^4}{8}\left((n+1)^4 - (n-1)^4\right) = \frac{n^4}{8}\left(8n^3 + 8n\right) = n^7 + n^5\end{align}$$ This clearly equals to ${\rm LHS}_n - {\rm LHS}_{n-1}$. As a result, $${\rm LHS}_n - {\rm RHS}_{n} = {\rm LHS}_{n-1} - {\rm RHS}_{n-1}$$ and the expression ${\rm LHS}_n - {\rm RHS}_{n}$ is independent of $n$. Since this difference vanishes at $n = 0$, we can conclude ${\rm LHS}_n = {\rm RHS}_n$ for all $n$ and hence establishes the identity at hand.

  • $\begingroup$ Simple and direct without any non-obvious tricks. +1 $\endgroup$
    – Paramanand Singh
    Dec 17 '17 at 4:43

Here are some interesting observations, which are too long to be included as a comment.

A useful reference can be found here.

Denoting $\displaystyle\sum_{r=1}^n r^m=\sigma_m$, we note that

$$\begin{align}\sigma_3&=\sigma_1^2\tag{1}\\ \frac{\sigma_5}{\sigma_3}&=\frac {4\sigma_1-1}{3}\tag{2}\\ \frac {\sigma_7}{\sigma_3}&=\frac {6\sigma_1^2-4\sigma_1+1}3\tag{3}\end{align}$$

Adding $(2),(3)$ and using $(1)$ gives $$\begin{align} \frac {\sigma_5+\sigma_7}{\sigma_3}&=2\sigma_1^2=2\sigma_3\\ \sigma_5+\sigma_7&=2\sigma_3^2=2\sigma_1^4\end{align}$$


Let the LHS be $A(n)$ and let the RHS be $2B(n)^4.$

Then $A(n+1)-A(n)=(n+1)^7+(n+1)^5=(n+1)^5(n^2+2n+2).$

$$\text {We have }\quad 2B(n+1)^4-2B(n)^4=$$ $$(*)\quad =2(B(n+1)^2+B(n)^2)\cdot (B(n+1)+B(n))\cdot (B(n+1)-B(n)).$$ Since $B(n)=n(n+1)/2$ we have $$B(n+1)^2+B(n)^2=(n+1)^2((n+2)^2+n^2)/4=(n+1)^2(n^2+2n+2)/2$$ $$\text { and}\quad B(n+1)+B(n)=(n+1)((n+2)+n)/2=(n+1)^2$$ $$\text { and }\quad B(n+1)-B(n)=n+1.$$ From these we compute $(*)$ (4th line from the top) and find it is equal to $(n+1)^5(n^2+2n+2)$, which is $A(n+1)-A(n).$ (2nd line from the top.).

So if $A(1)=B(1)$ then $A(n)=B(n)$ for all $n\in \Bbb N$ by induction on $n.$

Interesting identity. Another interesting one is $\sum_{x=1}^nx^3=(\sum_{x=1}^nx)^2=(n(n+1)/2)^2.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.