1. ## Interesting equality

I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you

Prove that for all natural n, the following holds:

$\displaystyle \left( {\sum\limits_{i = 1}^n i } \right)^2 = \sum\limits_{i = 1}^n {i^3 }$

For those who don't like summation symbols, the above is just math language to say that:

$\displaystyle \left( {1 + 2 + ... + n} \right)^2 = 1^3 + 2^3 + ... + n^3$

2. Originally Posted by TD!
I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you

Prove that for all natural n, the following holds:

$\displaystyle \left( {\sum\limits_{i = 1}^n i } \right)^2 = \sum\limits_{i = 1}^n {i^3 }$

For those who don't like summation symbols, the above is just math language to say that:

$\displaystyle \left( {1 + 2 + ... + n} \right)^2 = 1^3 + 2^3 + ... + n^3$

and the continuous analogue:

$\displaystyle \left(\int_0^x u\ du \right)^2=\int_0^x u^3\ du$,

which is easier to prove as we are more familiar with the rules for
manipulating integrals than we are in handling summations.

RonL

3. Indeed, but that one is rather easy to show

4. Originally Posted by TD!
Indeed, but that one is rather easy to show
but so is the other if you know the formula for the sum of cubes

RonL

5. Originally Posted by CaptainBlack
but so is the other if you know the formula for the sum of cubes

RonL
Well actually, that's what we're indirectly trying to prove so unfortunately, it's not allowed to use that formula

6. I admit that is an interesting equality, but it is rather easy to proof.

I have a different question find all integral solutions for $\displaystyle x,y$
such as,
$\displaystyle \sum^n_{k=1}k^x$=$\displaystyle (\sum^n_{k=1}k)^y$
for all $\displaystyle n\in Z^+$

For example $\displaystyle (x,y)=(3,2)$
But are there more?

7. Originally Posted by TD!
I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you

Prove that for all natural n, the following holds:

$\displaystyle \left( {\sum\limits_{i = 1}^n i } \right)^2 = \sum\limits_{i = 1}^n {i^3 }$

For those who don't like summation symbols, the above is just math language to say that:

$\displaystyle \left( {1 + 2 + ... + n} \right)^2 = 1^3 + 2^3 + ... + n^3$

we could use induction on the following:
$\displaystyle \left( {1 + 2 + ... + n} \right)^2 = 1^3 + 2^3 + ... + n^3$

8. Originally Posted by Jhevon
we could use induction on the following:
$\displaystyle \left( {1 + 2 + ... + n} \right)^2 = 1^3 + 2^3 + ... + n^3$
Or you can use addition formulas.

$\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$

$\displaystyle \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$.

I seen this identity like 15 times throughout my life.

9. Originally Posted by ThePerfectHacker
Or you can use addition formulas.

$\displaystyle \sum_{k=1}^n k = \frac{n(n+1)}{2}$

$\displaystyle \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$.

I seen this identity like 15 times throughout my life.
That's true, but i believe TD! said in an earlier post that we weren't allowed to use the formulas

Originally Posted by TD!
Well actually, that's what we're indirectly trying to prove so unfortunately, it's not allowed to use that formula

10. Induction would be a good approach, yes.

11. Originally Posted by TD!
Induction would be a good approach, yes.
were you thinking of something else?

$\displaystyle 2\left(\sum^n_1 k\right)^4\;=\;\sum^n_1 k^5 + \sum^n_1 k^7$

$\displaystyle 2(1 + 2 + 3 + \cdots + n)^4 \;=\;\left(1^5 + 2^5 + 3^5 + \hdots + n^5\right) + \left(1^7 + 2^7 + 3^7 + \hdots + n^7\right)$

The sum of the first $\displaystyle n$ positive integers, raised to the fourth power,
. . is the mean of the sum of the 5th powers and the sum of the 7th powers.

Let $\displaystyle T_n$ be the $\displaystyle n^{th}$ triangular number: .$\displaystyle T_n = \frac{n(n+1)}{2}$

$\displaystyle 3\left[\sum^n_1 T_k\right]^3 \;=\;\sum^n_1\left(T_k\right)^3 + 2\!\cdot\!\sum^n_1\left(T_k\right)^4$

$\displaystyle 3(1\ +\ 3\ +\ 6\ +\ \cdots\ +\ T_n)^3 \;=\;\left(1^3 + 3^3 + 6^3 + \cdots + T_n^3 \right)\,+\,2\left(1^4 + 3^4 + 6^4 + \cdots + T_n^4 \right)$

13. Originally Posted by Jhevon
were you thinking of something else?
No, induction was what I had in mind and what I used.

14. Originally Posted by ThePerfectHacker
I admit that is an interesting equality, but it is rather easy to proof.

I have a different question find all integral solutions for $\displaystyle x,y$
such as,
$\displaystyle \sum^n_{k=1}k^x$=$\displaystyle (\sum^n_{k=1}k)^y$
for all $\displaystyle n\in Z^+$

For example $\displaystyle (x,y)=(3,2)$
But are there more?
No

$\displaystyle \sum_{k=1}^n{k^r}\sim{\frac{n^{r+1}}{r+1}}$ for $\displaystyle r\geq{0}$ (asymptotically equal)

$\displaystyle {\text{As a necessary condition we have:}}\left( {\sum\limits_{k = 1}^n k } \right)^y \sim \sum\limits_{k = 1}^n {k^x }$

Since $\displaystyle \left( {\sum\limits_{k = 1}^n k } \right)^y \sim \frac{{n^{2y} }} {{2^y }},\sum\limits_{k = 1}^n {k^x } \sim \frac{{n^{x + 1} }} {{2^{x + 1} }}$

It is equivalent to $\displaystyle \left( {\sum\limits_{k = 1}^n k } \right)^y \sim \sum\limits_{k = 1}^n {k^x } \Leftrightarrow \frac{{n^{2y} }} {{2^y }} \sim \frac{{n^{x + 1} }} {{x + 1}}$

So it must be: $\displaystyle 2y = x + 1{\text{ and }}2^y = x + 1$

$\displaystyle \begin{gathered} {\text{It is easy to verify* that }}2^{y - 1} = y{\text{ has only 2 roots}} \hfill \\ y = 1,y = 2 \hfill \\ {\text{Thus }}\left( {x,y} \right) = (1,1){\text{ and }}\left( {x,y} \right) = (3,2){\text{ are the only}} \hfill \\ {\text{solutions}} \hfill \\ \end{gathered}$

* Derivatives

THe original problem
$\displaystyle \left( {\sum\limits_{k = 1}^n k } \right)^2 = \sum\limits_{k = 1}^n {k^2 } + 2 \cdot \sum\limits_{1 \leqslant k < j \leqslant n} {\left( {k \cdot j} \right)}$
$\displaystyle \sum\limits_{1 \leqslant k < j \leqslant n} {\left( {k \cdot j} \right)} = \sum\limits_{k = 1}^n {\left( {k \cdot \sum\limits_{j =0}^{k - 1} j } \right)} = \sum\limits_{k = 1}^n {\left( {k \cdot \tfrac{{k \cdot \left( {k - 1} \right)}} {2}} \right)}$
$\displaystyle \sum\limits_{1 \leqslant k < j \leqslant n} {\left( {k \cdot j} \right)} = \tfrac{1} {2} \cdot \sum\limits_{k = 1}^n {\left( {k^3 - k^2 } \right)} = \tfrac{1} {2} \cdot \sum\limits_{k = 1}^n {\left( {k^3 } \right)} - \tfrac{1} {2} \cdot \sum\limits_{k = 1}^n {\left( {k^2 } \right)}$
$\displaystyle \left( {\sum\limits_{k = 1}^n k } \right)^2 = \sum\limits_{k = 1}^n {k^2 } + 2 \cdot \left[ {\tfrac{1} {2} \cdot \sum\limits_{k = 1}^n {\left( {k^3 } \right)} - \tfrac{1} {2} \cdot \sum\limits_{k = 1}^n {\left( {k^2 } \right)} } \right] = \sum\limits_{k = 1}^n {\left( {k^3 } \right)} \blacksquare$

Actually I used the identity $\displaystyle \sum_{k=1}^n{k^2}=\frac{n\cdot{(n+1)}\cdot{(2n+1)} }{6}$ once in a math competition, but this one$\displaystyle \sum_{k=1}^n{k}=\frac{n\cdot{(n+1)}}{2}$ comes round all the time