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#### mente oscura

##### Well-known member

- Nov 29, 2013

- 172

[tex]Let \ T \in{N} \ / \ T=odd[/tex]

Prove:

[tex]1^3+3^3+5^3+7^3+ ... +T^3=\dfrac{[T(T+2)]^2-1}{8}[/tex]

Regards.

- Thread starter mente oscura
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- Thread starter
- #1

- Nov 29, 2013

- 172

[tex]Let \ T \in{N} \ / \ T=odd[/tex]

Prove:

[tex]1^3+3^3+5^3+7^3+ ... +T^3=\dfrac{[T(T+2)]^2-1}{8}[/tex]

Regards.

- Aug 18, 2013

- 76

$$=(1^3+2^3+3^3+\cdots +T^3)-(2^3+4^3+\cdots (T-1)^3)$$

$$=\frac{T(T+1)(2T+1)}{6}-8\Bigg(1^3+2^3+3^3+\cdots \bigg(\frac{T-1}{2}\bigg)^3\Bigg)$$

$$=\frac{T(T+1)(2T+1)}{6}-8\bigg(\frac{(\frac{T-1}{2})(\frac{T-1}{2}+1)(\frac{2T-2}{2}+1)}{6}\bigg)$$

Someone else can finish this off. I stopped trying.

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- #3

My solution:

\(\displaystyle S_n=\sum_{k=1}^n\left((2k-1)^3 \right)\)

This gives rise to the linear inhomogeneous difference equation:

\(\displaystyle S_n-S_{n-1}=(2n-1)^3\) where \(\displaystyle S_1=1\)

We can easily see that the homogenous solution is:

\(\displaystyle h_n=c_1\)

and the particular solution will thus take the form:

\(\displaystyle p_n=n\left(c_2+c_3n+c_4n^2+c_5n^3 \right)=c_2n+c_3n^2+c_4n^3+c_5n^4\)

Substituting the particular solution into the difference equation, we may solve for the parameters using the method of undetermined coefficients:

\(\displaystyle \left(c_2n+c_3n^2+c_4n^3+c_5n^4 \right)-\left(c_2(n-1)+c_3(n-1)^2+c_4(n-1)^3+c_5(n-1)^4 \right)=(2n-1)^3\)

\(\displaystyle 4c_5n^3+\left(3c_4-6c_5 \right)n^2+\left(2c_3-3c_4+4c_5 \right)n+\left(c_2-c_3+c_4-c_5 \right)=8n^3-12n^2+6n-1\)

Equating coefficients, we obtain the system:

\(\displaystyle 4c_5=8\)

\(\displaystyle 3c_4-6c_5=-12\)

\(\displaystyle 2c_3-3c_4+4c_5=6\)

\(\displaystyle c_2-c_3+c_4-c_5=-1\)

From this we obtain:

\(\displaystyle \left(c_2,c_3,c_4,c_5 \right)=(2,-1,0,0)\)

Hence, the particular solution is:

\(\displaystyle p_n=2n^4-n^2\)

Hence the general solution to the difference is:

\(\displaystyle S_n=h_n+p_n=c_1+2n^4-n^2\)

Using the initial value, we may determine the value of the parameter $c_1$:

\(\displaystyle S_1=c_1+2-1=c_1+1=1\implies c_1=0\)

Hence, the solution satisfying all conditions is:

\(\displaystyle S_n=2n^4-n^2=\frac{16n^4-8n^2+1-1}{8}=\frac{\left(4n^2-1 \right)^2-1}{8}=\frac{\left((2n-1)(2n+1) \right)^2-1}{8}\)

With $T\equiv2n-1$, we may state:

\(\displaystyle \sum_{k=1}^n\left((2n-1)^3 \right)=\frac{(T(T+2))^2-1}{8}\)

- Feb 15, 2012

- 1,967

With all due respect to MarkFL, the simplest proof is

Let $T = 2n + 1$. We must then show for all $n \in \Bbb N$:

$\displaystyle \sum_{k = 0}^n (2k + 1)^3 = \frac{[(2n + 1)(2n + 3)]^2 - 1}{8}$

Clearly, we have, for $n = 0$:

$1^3 = \dfrac{[(1)(3)]^2 - 1}{8} = \dfrac{9 - 1}{8} = \dfrac{8}{8} = 1$.

Assume that for $n = m$, we have:

$\displaystyle \sum_{k = 0}^m (2k + 1)^3 = \frac{[(2m + 1)(2m + 3)]^2 - 1}{8}$

Then:

$\displaystyle \sum_{k = 0}^{m+1} (2k + 1)^3 = \left(\sum_{k = 0}^m (2k + 1)^3\right) + (2m + 3)^3$

$\displaystyle = \frac{[(2m + 1)(2m + 3)]^2 - 1}{8} + \frac{8(2m + 3)^3}{8}$

$\displaystyle = \frac{(2m + 3)^2[(2m + 1)^2 + 8(2m + 3)] - 1}{8}$

$\displaystyle = \frac{(2m + 3)^2[4m^2 + 4m + 1 + 16m + 24] - 1}{8}$

$\displaystyle = \frac{(2m + 3)^2[4m^2 + 20m + 25] - 1}{8}$

$\displaystyle = \frac{[(2m + 3)(2m + 5)]^2 - 1}{8}$

- Thread starter
- #5

- Nov 29, 2013

- 172

Hello.

A picturesque prove:

[tex]3^3=2+3+4+5+6+7[/tex]

[tex]5^3=8+9+10+11+12+13+14+15+16+17[/tex]

[tex]7^3=18+19+20+21+22+23+24+25+26+27+28+29+30+31[/tex]

...

...

[tex]T^3=\dfrac{T^2-2T+1}{2}+ ... +\dfrac{T^2+2T-1}{2}[/tex]

It would then be an arithmetic progression.

First term=[tex]\ 1[/tex]

Last term=[tex]\dfrac{T^2+2T-1}{2}[/tex]

Total number of terms=[tex]\dfrac{T^2+2T-1}{2}[/tex]

Solution:

[tex]1^3+3^3+...+T^3=\dfrac{(1)+(\dfrac{T^2+2T-1}{2})}{2} \ \dfrac{T^2+2T-1}{2}=[/tex]

[tex]=\dfrac{\dfrac{T^2+2T+1}{2}}{2} \ \dfrac{T^2+2T-1}{2}=[/tex]

[tex]=\dfrac{[T(T+2)]^2-1}{8}[/tex]

Regards.

- Feb 15, 2012

- 1,967

As with many such proofs, I found your opening statements the most difficult to justify to myself. However:Hello.

A picturesque prove:

[tex]3^3=2+3+4+5+6+7[/tex]

[tex]5^3=8+9+10+11+12+13+14+15+16+17[/tex]

[tex]7^3=18+19+20+21+22+23+24+25+26+27+28+29+30+31[/tex]

...

...

[tex]T^3=\dfrac{T^2-2T+1}{2}+ ... +\dfrac{T^2+2T-1}{2}[/tex]

It would then be an arithmetic progression.

First term=[tex]\ 1[/tex]

Last term=[tex]\dfrac{T^2+2T-1}{2}[/tex]

Total number of terms=[tex]\dfrac{T^2+2T-1}{2}[/tex]

Solution:

[tex]1^3+3^3+...+T^3=\dfrac{(1)+(\dfrac{T^2+2T-1}{2})}{2} \ \dfrac{T^2+2T-1}{2}=[/tex]

[tex]=\dfrac{\dfrac{T^2+2T+1}{2}}{2} \ \dfrac{T^2+2T-1}{2}=[/tex]

[tex]=\dfrac{[T(T+2)]^2-1}{8}[/tex]

Regards.

\(\displaystyle = 2T\left(\frac{T^2 - 2T + 1}{2}\right) + \sum_{k=1}^{2T} k-1\)

\(\displaystyle = T(T-1)^2 + \sum_{j=1}^{2T-1}j\)

\(\displaystyle = T(T-1)^2 + \frac{(2T-1)(2T)}{2}\)

\(\displaystyle = T(T^2 - 2T + 1) + T(2T - 1) = T^3\)

as you indeed claim, so I am satisfied

- Admin
- #7

Yes, being given the closed form, induction is indeed simpler and more straightforward. However, I wanted to derive the result instead.With all due respect to MarkFL, the simplest proof isQED.

Let $T = 2n + 1$. We must then show for all $n \in \Bbb N$:

$\displaystyle \sum_{k = 0}^n (2k + 1)^3 = \frac{[(2n + 1)(2n + 3)]^2 - 1}{8}$

Clearly, we have, for $n = 0$:

$1^3 = \dfrac{[(1)(3)]^2 - 1}{8} = \dfrac{9 - 1}{8} = \dfrac{8}{8} = 1$.

Assume that for $n = m$, we have:

$\displaystyle \sum_{k = 0}^m (2k + 1)^3 = \frac{[(2m + 1)(2m + 3)]^2 - 1}{8}$

Then:

$\displaystyle \sum_{k = 0}^{m+1} (2k + 1)^3 = \left(\sum_{k = 0}^m (2k + 1)^3\right) + (2m + 3)^3$

$\displaystyle = \frac{[(2m + 1)(2m + 3)]^2 - 1}{8} + \frac{8(2m + 3)^3}{8}$

$\displaystyle = \frac{(2m + 3)^2[(2m + 1)^2 + 8(2m + 3)] - 1}{8}$

$\displaystyle = \frac{(2m + 3)^2[4m^2 + 4m + 1 + 16m + 24] - 1}{8}$

$\displaystyle = \frac{(2m + 3)^2[4m^2 + 20m + 25] - 1}{8}$

$\displaystyle = \frac{[(2m + 3)(2m + 5)]^2 - 1}{8}$

- Jan 17, 2013

- 1,667

Mostly a derivation is a little bit more satisfactory than a proof by induction. Well, at least for me

- Feb 15, 2012

- 1,967

This is understandable...most inductive proofs come about by "already knowing the answer", which begs the question: "how did you find the answer in the FIRST place?"Mostly a derivation is a little bit more satisfactory than a proof by induction. Well, at least for me

Of course, one might make an "inductive leap" by observing what appears to be a pattern, but in many cases, the complexity grows so fast that it appears impossible to use such an approach "all the time".

I find mente oscura's proof a "middle ground", he does not use induction per se (although it is certainly lurking in the background with the closed form for an arithmetic progression), but it is still presented on more or less "elementary terms" using only basic algebraic tools.

That said, it is often the case that "sophisticated methods" are powerful tools for investigating structures much simpler than the more complicated settings they arise in. The better one is versed in such methods, the wider array of problems one is able to tackle with them.

One final note:

The theory of difference equations depends in an essential way on the ability to recursively define functions, the soundness of such an approach is actually equivalent to the principle of induction (without which no recursive definition would even be possible). Usually this is way,way in the background, and taken as given.

I myself am more fond of the "idiot's approach", perhaps a sign of my own limited imagination...certainly MarkFL's proof is impressive in its scope, and one I would not have thought of until I saw it. It has an undeniably "analytic" flavor, leading me to believe he is more at home with analytic methods than algebraic ones, and perhaps Zaid's enthusiasm for his proof indicates a similar propensity.

(When I took analysis, I enjoyed the theorems, but I simply *detested* the exercises, which seemed to me to be boring chores in symbolic manipulation).

- Jan 31, 2012

- 253

Another approach is to use the Euler-Macluarin summation formula.

$$\sum_{k=1}^{n} (2n-1)^{3} = \sum_{k=0}^{n-1} (2n+1)^{3} = \int_{0}^{n} (2x+1)^{3} \ dx + B_{1} \Big((2n+1)^3 - 1 \Big) + \frac{B_{2}}{2!} \Big(6(2n+1)^2-6\Big) $$

$$ = \frac{1}{8} \Big((2n+1)^{4} -1 \Big) - \frac{1}{2} \Big((2n+1)^{3}-1 \Big) + \frac{1}{2} \Big((2n+1)^{2}-1 \Big)$$

$$ = \frac{(2n+1)^{4} - 4 (2n+1)^{3}+ 4(2n+1)^{2}-1}{8}$$

$$ = \frac{(2n+1)^{2}\Big( (2n+1)^{2}-4(2n+1)+4 \Big)-1}{8} = \frac{(2n+1)^{2}(2n-1)^{2}-1}{8}$$

Last edited:

- Mar 22, 2013

- 573

Best answer of all, RV. You're a genius. Period.

- Thread starter
- #12

- Nov 29, 2013

- 172

Hello.

I have opened a thread in which I expose a few formulas, in which I I've based, for this issue.

In Number Theory.

Regards.

Using this, notice that

[tex]\displaystyle \begin{align*} 2^3 + 4^3 + 6^3 + \dots + \left( 2n \right) ^3 &= \left( 2 \cdot 1 \right) ^3 + \left( 2 \cdot 2 \right) ^3 + \left( 2 \cdot 3 \right) ^3 + \dots + \left( 2 \cdot n \right) ^3 \\ &= 2^3 \cdot 1^3 + 2^3 \cdot 2^3 + 2^3 \cdot 3^3 + \dots + 2^3 \cdot n^3 \\ &= 2^3 \left( 1^3 + 2^3 + 3^3 + \dots + n^3 \right) \\ &= 8 \left\{ \frac{1}{4} \left[ n \left( n + 1 \right) \right]^2 \right\} \\ &= 2 \left[ n \left( n + 1 \right) \right] ^2 \end{align*}[/tex]

and thus, with [tex]\displaystyle \begin{align*} T = 2n + 1 \end{align*}[/tex], we have

[tex]\displaystyle \begin{align*} 1^3 + 3^3 + 5^3 + \dots + T^3 &= 1^3 + 3^3 + 5^3 + \dots + \left( 2n + 1 \right) ^3 \\ &= \left[ 1^3 + 2^3 + 3^3 + \dots + \left( 2n \right) ^3 + \left( 2n + 1 \right) ^3 \right] - \left[ 2^3 + 4^3 + 6^3 + \dots + \left( 2n \right) ^3 \right] \\ &= \frac{1}{4} \left[ \left( 2n + 1 \right) \left( 2n + 1 + 1 \right) \right] ^2 - 2 \left[ n \left( n + 1 \right) \right] ^2 \\ &= \left( 2n + 1 \right) ^2 \left( n + 1 \right) ^2 - 2n^2 \left( n + 1 \right) ^2 \\ &= \left( n + 1 \right) ^2 \left[ \left( 2n + 1 \right) ^2 - 2n^2 \right] \\ &= \left( \frac{T + 1}{2} \right) ^2 \left[ T^2 - \frac{\left( T - 1 \right) ^2 }{2} \right] \\ &= \frac{\left( T + 1 \right) ^2}{4} \left[ \frac{T^2 + 2T - 1}{2} \right] \\ &= \frac{\left( T + 1 \right) ^2 \left[ \left( T + 1 \right) ^2 - 2 \right] }{8} \\ &= \frac{ \left( T + 1 \right) ^4 - 2 \left( T + 1 \right) ^2 }{8} \\ &= \frac{ T^4 + 4T^3 + 6T^2 + 4T + 1 - 2T^2 - 4T - 2 }{8} \\ &= \frac{T^4 + 4T^3 + 4T^2 - 1}{8} \\ &= \frac{T^2 \left( T^2 + 4T + 4 \right) - 1}{8} \\ &= \frac{T^2 \left( T + 2 \right) ^2 - 1}{8} \\ &= \frac{\left[ T \left( T + 2 \right) \right] ^2 - 1}{8} \end{align*}[/tex]

Q.E.D.

- Jan 17, 2013

- 1,667

At first glance when I saw the question I thought about induction. Usually I don't like the discrete and I am in favor of continuity. But as I am majoring in computer science it is becoming a must that I employ the discrete methods. The transformation from a discrete quantity to a continuous quantity and vice versa is usually an interesting question. Winding numbers in algebraic topology is an example of such interesting transformations.This is understandable...most inductive proofs come about by "already knowing the answer", which begs the question: "how did you find the answer in the FIRST place?"

Of course, one might make an "inductive leap" by observing what appears to be a pattern, but in many cases, the complexity grows so fast that it appears impossible to use such an approach "all the time".

I find mente oscura's proof a "middle ground", he does not use induction per se (although it is certainly lurking in the background with the closed form for an arithmetic progression), but it is still presented on more or less "elementary terms" using only basic algebraic tools.

That said, it is often the case that "sophisticated methods" are powerful tools for investigating structures much simpler than the more complicated settings they arise in. The better one is versed in such methods, the wider array of problems one is able to tackle with them.

One final note:

The theory of difference equations depends in an essential way on the ability to recursively define functions, the soundness of such an approach is actually equivalent to the principle of induction (without which no recursive definition would even be possible). Usually this is way,way in the background, and taken as given.

I myself am more fond of the "idiot's approach", perhaps a sign of my own limited imagination...certainly MarkFL's proof is impressive in its scope, and one I would not have thought of until I saw it. It has an undeniably "analytic" flavor, leading me to believe he is more at home with analytic methods than algebraic ones, and perhaps Zaid's enthusiasm for his proof indicates a similar propensity.

(When I took analysis, I enjoyed the theorems, but I simply *detested* the exercises, which seemed to me to be boring chores in symbolic manipulation).

PS: Sorry for the OP for the off-topic discussion but usually a proof by induction carries lots of discussions afterwards .

- Admin
- #15

A bit of justification on the solution I posted:

\(\displaystyle S_n=An^4+Bn^3+Cn^2+Dn+E\)

This being the superposition of the homogenous and particular solutions. Recall I began with the linear difference equation (which I will arrange recursively):

\(\displaystyle S_n=S_{n-1}+(2n-1)^3\)

Let's expand the binomial:

(1) \(\displaystyle S_{n}=S_{n-1}+8n^3-12n^2+6n-1\)

We may of course also write the equivalent:

(2) \(\displaystyle S_{n+1}=S_{n}+8(n+1)^3-12(n+1)^2+6(n+1)-1\)

Subtracting (1) from (2) we obtain:

(3) \(\displaystyle S_{n+1}=2S_{n}-S_{n-1}+24n^2+2\)

(4) \(\displaystyle S_{n+2}=2S_{n+1}-S_{n}+24(n+1)^2+2\)

Subtracting (3) from (4) we obtain:

(5) \(\displaystyle S_{n+2}=3S_{n+1}-3S_{n}+S_{n-1}+48n+24\)

(6) \(\displaystyle S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+48(n+1)+24\)

Subtracting (5) from (6) we obtain:

(7) \(\displaystyle S_{n+3}=4S_{n+2}-6S_{n+1}+4S_{n}-S_{n-1}+48\)

(8) \(\displaystyle S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}+48\)

Subtracting (7) from (8) we obtain:

\(\displaystyle S_{n+4}=5S_{n+3}-10S_{n+2}+10S_{n+1}-5S_{n}+S_{n-1}\)

From this method of symbolic differencing, we now have a homogenous recursion whose associated characteristic equation is:

\(\displaystyle r^5-5r^4+10r^3-10r^2+5r-1=(r-1)^5=0\)

Since we have the characteristic root $r=1$ of multiplicity 5, we know the general solution must have the form:

\(\displaystyle S_n=An^4+Bn^3+Cn^2+Dn+E\)

Upon further reflection and reading of this thread, I feel my post is incomplete, as the picture I posted, while beautiful, only suggests a patterns and shows its statement is true for the first five natural numbers. Although the pattern is obvious, I don't feel it is proved...

If we are trying to prove [tex]\displaystyle \begin{align*} 1^3 + 2^3 + 3^3 + \dots + n^3 = \frac{1}{4} \left[ n \left( n + 1 \right) \right] ^2 \end{align*}[/tex] then Base Step ([tex]\displaystyle \begin{align*} n = 1 \end{align*}[/tex]):

[tex]\displaystyle \begin{align*} LHS &= 1^3 \\ &= 1 \\ \\ RHS &= \frac{1}{4} \left[ 1 \cdot \left( 1 + 1 \right) \right] ^2 \\ &= \frac{1}{4} \cdot 2^2 \\ &= 1 \\ &= LHS \end{align*}[/tex]

Assume the statement is true for [tex]\displaystyle \begin{align*} n = k \end{align*}[/tex], i.e. [tex]\displaystyle \begin{align*} 1^3 + 2^3 + 3^3 + \dots + k^3 &= \frac{1}{4} \left[ k \left( k + 1 \right) \right] ^2 \end{align*}[/tex]

Now show the statement is true for [tex]\displaystyle \begin{align*} n = k + 1 \end{align*}[/tex], i.e. show [tex]\displaystyle \begin{align*} 1^3 + 2^3 + 3^3 + \dots + k^3 + \left( k + 1 \right) ^3 &= \frac{1}{4} \left[ \left( k + 1 \right) \left( k + 2 \right) \right] ^2 \end{align*}[/tex]

Inductive Step:

[tex]\displaystyle \begin{align*} 1^3 + 2^3 + 3^3 + \dots + k^3 + \left( k + 1 \right) ^3 &= \frac{1}{4} \left[ k \left( k + 1 \right) \right] ^2 + \left( k + 1 \right) ^3 \\ &= \frac{1}{4} k^2 \left( k + 1 \right) ^2 + \left( k + 1 \right) ^3 \\ &= \left( k + 1 \right) ^2 \left[ \frac{1}{4}k^2 + k + 1 \right] \\ &= \frac{1}{4} \left( k + 1 \right) ^2 \left( k^2 + 4k + 4 \right) \\ &= \frac{1}{4} \left( k + 1 \right) ^2 \left( k + 2 \right) ^2 \end{align*}[/tex]

Q.E.D.

- Feb 15, 2012

- 1,967

I've always found it fascinating that:

$$\sum_{k=1}^n k^3 = \left[\sum_{k=1}^n k\right]^2$$

- Mar 22, 2013

- 573

Do you think this is fascinating too :

$$\sum_{d|n} \tau(d)^3 = \left [ \sum_{d|n} \tau(d) \right ]^2$$

or

$$\sum_{k = 1}^n n^3 = \left [ \sum_{k = 1}^n n \right ]^2$$

And if you do, how about the multisets

$$\{1, 2, 2, 3, 5\}, \{3, 3, 3, 3, 6\}, \text{etc}$$

with the same property but in neither of the three classes, including the one you've given above. Also, [Hold you chair and don't freak out!]

$$\bigg \{\frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, \frac23, 2, 2, 10 \bigg\}$$

$$\{1 + 11678679130451250771390i, 10114032809617941274227 + 5839339565225625385695i\}$$

They have some strange properties, mostly of number theoretic interest, although there might be something for you here :

$$\{1, 2, 3\} \otimes \{1, 2\} = \{1, 2, 2, 3, 4, 6\}$$

This operation forms a semigroup.

PS : All of these are recorded in MMF in 5 topics : Diophantine Equation with Multiple Variables, Elements Of The J-Semigroup, J Semigroup, J Semigroup Again, J commutative monoid [read by order].