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Consider the following:

$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$

$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$

$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$

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

Is

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

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    $\begingroup$ Could you prove this by induction? $\endgroup$ Dec 15 '17 at 17:42
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ Simple and direct without any non-obvious tricks. +1 $\endgroup$
    – Paramanand Singh
    Dec 17 '17 at 4:43
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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 $$\begin{align} \frac {\sigma_5+\sigma_7}{\sigma_3}&=2\sigma_1^2=2\sigma_3\\ \sigma_5+\sigma_7&=2\sigma_3^2=2\sigma_1^4\end{align}$$

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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 $\sum_{x=1}^nx^3=(\sum_{x=1}^nx)^2=(n(n+1)/2)^2.$

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