Interesting equality

**TD!** · January 13th 2006, 01:14 PM

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:

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

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

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

Please don't reveal your answer yet, or post it 'covered'.

**CaptainBlack** · January 13th 2006, 01:27 PM

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:

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

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

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

Please don't reveal your answer yet, or post it 'covered'.

and the continuous analogue:

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

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

RonL

**TD!** · January 13th 2006, 01:30 PM

Indeed, but that one is rather easy to show

**CaptainBlack** · January 13th 2006, 01:34 PM

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

**TD!** · January 14th 2006, 09:26 AM

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

**ThePerfectHacker** · January 14th 2006, 02:33 PM

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

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

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

**Jhevon** · May 24th 2007, 05:04 PM

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:

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

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

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

Please don't reveal your answer yet, or post it 'covered'.

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

**ThePerfectHacker** · May 24th 2007, 06:23 PM

Originally Posted by Jhevon

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

Or you can use addition formulas.

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

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

I seen this identity like 15 times throughout my life.

**Jhevon** · May 24th 2007, 06:52 PM

Originally Posted by ThePerfectHacker

Or you can use addition formulas.

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

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

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

**TD!** · May 26th 2007, 06:06 AM

Induction would be a good approach, yes.

**Jhevon** · May 26th 2007, 07:10 AM

Originally Posted by TD!

Induction would be a good approach, yes.

were you thinking of something else?

**Soroban** · May 26th 2007, 10:02 AM

How about these identities?

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

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

The sum of the first $n$ 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 $T_n$ be the $n^{th}$ triangular number: . $T_n = \frac{n(n+1)}{2}$

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

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

**TD!** · May 26th 2007, 11:35 AM

Originally Posted by Jhevon

were you thinking of something else?

No, induction was what I had in mind and what I used.

**PaulRS** · January 4th 2008, 05:57 AM

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 $x,y$
such as,
$\sum^n_{k=1}k^x$ = $(\sum^n_{k=1}k)^y$
for all $n\in Z^+$

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

No

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

$ {\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 $ \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 $ \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: $ 2y = x + 1{\text{ and }}2^y = x + 1 $

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

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