Showing that the sum of cubes of an arithmetic progression is divisible by the sum of its terms

[Aion]

Find the sum of the cubes of the terms of an arithmetic progression and show that it is exactly divisible by the sum of the terms.

Attempt

Consider the arithmetic progression
[math]a + kd,\quad k = 0,1,\dots,n-1[/math]
The sum of the terms is
[math]S = \sum_{k=0}^{n-1}(a+kd) = \frac{n}{2}\big(2a+(n-1)d\big)[/math]
We consider the sum of cubes
[math]T = \sum_{k=0}^{n-1}(a+kd)^3[/math]
Expanding,
[math]T = \sum_{k=0}^{n-1}\left(a^3 + 3a^2kd + 3ad^2k^2 + d^3k^3\right)[/math]
so
[math]T = na^3 + 3a^2d\sum_{k=0}^{n-1}k + 3ad^2\sum_{k=0}^{n-1}k^2 + d^3\sum_{k=0}^{n-1}k^3[/math]
Using standard formulas,
[math]\sum_{k=0}^{n-1} k = \frac{n(n-1)}{2},\quad \sum_{k=0}^{n-1} k^2 = \frac{n(n-1)(2n-1)}{6},\quad \sum_{k=0}^{n-1} k^3 = \left(\frac{n(n-1)}{2}\right)^2[/math],

we obtain
[math]T = na^3 + \frac{3}{2}a^2d\,n(n-1) + \frac{1}{2}ad^2\,n(n-1)(2n-1) + \frac{d^3}{4}n^2(n-1)^2[/math]
Factoring out n/4,
[math]T =\frac{n}{4}\Big(4a^3+ 6a^2d(n-1)+ 2ad^2(n-1)(2n-1)+ d^3n(n-1)^2\Big)[/math]
At this stage, I tried to factor out the known sum
[math]S = \frac{n}{2}\big(2a+(n-1)d\big)[/math]but a direct algebraic factorisation from the fully expanded form was not apparent. I also tried introducing the substitution [imath]x = (n-1)d[/imath] to rewrite the numerator in the form [imath](2a+x)(\text{quadratic in} \, a,x)[/imath], but this did not lead to a clean factorisation since the substitution mixes powers of [imath]d[/imath] in a way that obscures the structure.

I am also aware of an alternative approach based on symmetry of arithmetic progressions, which I include below as it provides a more conceptual explanation of why the factorisation works.

Alternative approach (symmetry).

Let
[math]x_k = a + kd[/math]
Write this in centred form using
[math]m = \frac{2a + (n-1)d}{2},\quad u_k = k - \frac{n-1}{2}[/math]so that
[math]x_k = m + du_k[/math]
Then
[math]\sum x_k^3 = \sum (m + du_k)^3 = nm^3 + 3m^2d\sum u_k + 3md^2\sum u_k^2 + d^3\sum u_k^3[/math]
Since the sequence [imath]u_k[/imath] is symmetric about [imath]0[/imath],
[math]\sum u_k = 0,\quad \sum u_k^3 = 0[/math]
so this reduces to
[math]\sum x_k^3 = nm^3 + 3md^2\sum u_k^2[/math]
Using
[math]\sum_{k=0}^{n-1}\left(k-\frac{n-1}{2}\right)^2 = \frac{n(n^2-1)}{12}[/math]
we get
[math]\sum x_k^3 = nm\left(m^2 + d^2\frac{n^2-1}{4}\right)[/math]
Since [imath]S = nm[/imath], it follows that
[math]\sum x_k^3 = S\left(m^2 + d^2\frac{n^2-1}{4}\right)[/math]
which shows that the sum of cubes is divisible by the sum of terms.

 

[lookagain]

Find the sum of the cubes of the terms of an arithmetic progression and show that it is exactly divisible by the sum of the terms.

Your topic problem (and demonstration of the attempt of the answer) belong under the forum "Intermediate/Advanced Algebra."

 

[Aion]

Your topic problem (and demonstration of the attempt of the answer) belong under the forum "Intermediate/Advanced Algebra."

Thank you for letting me know. I wasn’t entirely sure which forum category was most appropriate for the problem.

 

[Aion]

This approach resolves the earlier difficulty I had with directly factoring the expanded expression. In particular, instead of trying to manipulate the full polynomial form of [imath]T[/imath], the key idea is to exploit the fact that the expression contains a known linear factor

[math]S = \frac{n}{2}\bigl(2a+(n-1)d\bigr)[/math]
This shifts the problem away from brute-force algebra and toward understanding what remains after factoring out [imath]S[/imath]. Once this is done, the remaining factor can be determined systematically.

We have already shown that
[math]T=na^3+\frac{3}{2}a^2d\,n(n-1)+\frac{1}{2}ad^2\,n(n-1)(2n-1)+\frac{d^3}{4}n^2(n-1)^2[/math]
Since
[math]S=\frac{n}{2}\bigl(2a+(n-1)d\bigr)[/math]is expected to be a factor of [imath]T[/imath], we write
[math]T=S\cdot Q(a,d)[/math]where [imath]Q(a,d)[/imath] must be homogeneous of degree [imath]2[/imath] in [imath]a[/imath] and [imath]d[/imath]

Thus we assume
[math]Q(a,d)=\alpha a^2+\beta ad+\gamma d^2[/math]where [imath]\alpha,\beta,\gamma[/imath] may depend on [imath]n[/imath]

Substituting [imath]S[/imath], we obtain
[math]T = \frac{n}{2}\bigl(2a+(n-1)d\bigr) \bigl(\alpha a^2+\beta ad+\gamma d^2\bigr)[/math]
Expanding symbolically
[math]\begin{aligned} T &= \frac{n}{2}\Bigl[2a(\alpha a^2+\beta ad+\gamma d^2)+(n-1)d(\alpha a^2+\beta ad+\gamma d^2)\Bigr] \\[1ex] &= \frac{n}{2}\Bigl[2\alpha a^3+2\beta a^2d+2\gamma ad^2+(n-1)\alpha a^2d+(n-1)\beta ad^2+(n-1)\gamma d^3\Bigr] \\[1ex] &= \frac{n}{2}\Bigl[2\alpha a^3+\bigl(2\beta+(n-1)\alpha\bigr)a^2d+\bigl(2\gamma+(n-1)\beta\bigr)ad^2+(n-1)\gamma d^3\Bigr] \end{aligned}[/math]
Now compare coefficients with the known expansion of [imath]T[/imath].

For the [imath]a^3[/imath]-term:
[math]\frac{n}{2}(2\alpha)=n[/math]hence
[math]\alpha=1[/math]
For the [imath]a^2d[/imath]-term:
[math]\frac{n}{2}\bigl(2\beta+(n-1)\bigr) = \frac{3}{2}n(n-1)[/math]
Cancelling [imath]n/2[/imath]
[math]2\beta+(n-1)=3(n-1)[/math]so
[math]\beta=n-1[/math]
For the [imath]ad^2[/imath]-term:
[math]\frac{n}{2}\bigl(2\gamma+(n-1)^2\bigr) = \frac{1}{2}n(n-1)(2n-1)[/math]
Cancelling [imath]n/2[/imath]
[math]2\gamma+(n-1)^2=(n-1)(2n-1)[/math]
Therefore
[math]2\gamma = (n-1)\bigl((2n-1)-(n-1)\bigr) =n(n-1)[/math]and hence
[math]\gamma=\frac{n(n-1)}{2}[/math]
Thus
[math]Q(a,d) = a^2+(n-1)ad+\frac{n(n-1)}{2}d^2[/math]
Substituting back
[math]T=\frac{n}{2}\bigl(2a+(n-1)d\bigr) \left(a^2+(n-1)ad+\frac{n(n-1)}{2}d^2\right)[/math]

 

[nasi112]

Why you did not use the direct evidence?

[math]\frac{T}{S} = \frac{\sum_{k=0}^{n-1}(a + kd)^3}{\sum_{k=0}^{n-1}(a + kd)}[/math]
[math]= \sum_{k=0}^{n-1}\frac{(a + kd)^3}{(a + kd)}[/math]
[math]= \sum_{k=0}^{n-1}(a + kd)^2[/math]

 

[Aion]

Why you did not use the direct evidence?

[math]\frac{T}{S} = \frac{\sum_{k=0}^{n-1}(a + kd)^3}{\sum_{k=0}^{n-1}(a + kd)}[/math]
[math]= \sum_{k=0}^{n-1}\frac{(a + kd)^3}{(a + kd)}[/math]
[math]= \sum_{k=0}^{n-1}(a + kd)^2[/math]

The step
[math]\frac{\sum_{k=0}^{n-1}(a+kd)^3}{\sum_{k=0}^{n-1}(a+kd)} = \sum_{k=0}^{n-1}(a+kd)^2[/math]is not valid in general. One cannot “cancel” terms inside sums in this way, since
[math]\frac{\sum a_k}{\sum b_k} \neq \sum \frac{a_k}{b_k}.[/math]
A simple counterexample is [imath]a=1[/imath], [imath]d=1[/imath], [imath]n=2[/imath]. Then
[math]\frac{1^3+2^3}{1+2}=\frac{9}{3}=3[/math]whereas
[math]1^2+2^2=5[/math]
Hence the proposed simplification does not hold.

 

[nasi112]

I am speechless after seeing your counterexample.

Good job Aion.

I need to sharpen my algebra skills.

 

[scottbrandon]

Find the sum of the cubes of the terms of an arithmetic progression and show that it is exactly divisible by the sum of the terms.

Attempt

Consider the arithmetic progression
[math]a + kd,\quad k = 0,1,\dots,n-1[/math]
The sum of the terms is
[math]S = \sum_{k=0}^{n-1}(a+kd) = \frac{n}{2}\big(2a+(n-1)d\big)[/math]
We consider the sum of cubes
[math]T = \sum_{k=0}^{n-1}(a+kd)^3[/math]
Expanding,
[math]T = \sum_{k=0}^{n-1}\left(a^3 + 3a^2kd + 3ad^2k^2 + d^3k^3\right)[/math]
so
[math]T = na^3 + 3a^2d\sum_{k=0}^{n-1}k + 3ad^2\sum_{k=0}^{n-1}k^2 + d^3\sum_{k=0}^{n-1}k^3[/math]
Using standard formulas,
[math]\sum_{k=0}^{n-1} k = \frac{n(n-1)}{2},\quad \sum_{k=0}^{n-1} k^2 = \frac{n(n-1)(2n-1)}{6},\quad \sum_{k=0}^{n-1} k^3 = \left(\frac{n(n-1)}{2}\right)^2[/math],

we obtain
[math]T = na^3 + \frac{3}{2}a^2d\,n(n-1) + \frac{1}{2}ad^2\,n(n-1)(2n-1) + \frac{d^3}{4}n^2(n-1)^2[/math]
Factoring out n/4,
[math]T =\frac{n}{4}\Big(4a^3+ 6a^2d(n-1)+ 2ad^2(n-1)(2n-1)+ d^3n(n-1)^2\Big)[/math]
At this stage, I tried to factor out the known sum
[math]S = \frac{n}{2}\big(2a+(n-1)d\big)[/math]but a direct algebraic factorisation from the fully expanded form was not apparent. I also tried introducing the substitution [imath]x = (n-1)d[/imath] to rewrite the numerator in the form [imath](2a+x)(\text{quadratic in} \, a,x)[/imath], but this did not lead to a clean factorisation since the substitution mixes powers of [imath]d[/imath] in a way that obscures the structure.

I am also aware of an alternative approach based on symmetry of arithmetic progressions, which I include below, much like a ps2 emulator for android, as it provides a more conceptual explanation of why the factorisation works.

Alternative approach (symmetry).

Let
[math]x_k = a + kd[/math]
Write this in centred form using
[math]m = \frac{2a + (n-1)d}{2},\quad u_k = k - \frac{n-1}{2}[/math]so that
[math]x_k = m + du_k[/math]
Then
[math]\sum x_k^3 = \sum (m + du_k)^3 = nm^3 + 3m^2d\sum u_k + 3md^2\sum u_k^2 + d^3\sum u_k^3[/math]
Since the sequence [imath]u_k[/imath] is symmetric about [imath]0[/imath],
[math]\sum u_k = 0,\quad \sum u_k^3 = 0[/math]
so this reduces to
[math]\sum x_k^3 = nm^3 + 3md^2\sum u_k^2[/math]
Using
[math]\sum_{k=0}^{n-1}\left(k-\frac{n-1}{2}\right)^2 = \frac{n(n^2-1)}{12}[/math]
we get
[math]\sum x_k^3 = nm\left(m^2 + d^2\frac{n^2-1}{4}\right)[/math]
Since [imath]S = nm[/imath], it follows that
[math]\sum x_k^3 = S\left(m^2 + d^2\frac{n^2-1}{4}\right)[/math]
which shows that the sum of cubes is divisible by the sum of terms.

Your symmetry approach is the cleanest proof. Let x_k = a + kd, where k = 0, 1, ..., n − 1. Define m = [2a + (n − 1)d]/2 and u_k = k − (n − 1)/2, so x_k = m + du_k. The u_k values are symmetric about zero, hence sum u_k = 0 and sum u_k^3 = 0. Therefore, sum(x_k^3) = nm^3 + 3md^2 sum(u_k^2). Since sum(u_k^2) = n(n^2 − 1)/12, we get sum(x_k^3) = nm[m^2 + d^2(n^2 − 1)/4]. But the sum of the original arithmetic progression is S = nm. Hence sum(x_k^3) = S[m^2 + d^2(n^2 − 1)/4], which proves that the sum of the cubes is exactly divisible by the sum of the terms.

 

[blamocur]

Your symmetry approach is the cleanest proof. Let x_k = a + kd, where k = 0, 1, ..., n − 1. Define m = [2a + (n − 1)d]/2 and u_k = k − (n − 1)/2, so x_k = m + du_k. The u_k values are symmetric about zero, hence sum u_k = 0 and sum u_k^3 = 0. Therefore, sum(x_k^3) = nm^3 + 3md^2 sum(u_k^2). Since sum(u_k^2) = n(n^2 − 1)/12, we get sum(x_k^3) = nm[m^2 + d^2(n^2 − 1)/4]. But the sum of the original arithmetic progression is S = nm. Hence sum(x_k^3) = S[m^2 + d^2(n^2 − 1)/4], which proves that the sum of the cubes is exactly divisible by the sum of the terms.

Instead of contributing to old threads you could spend time on learning KaTeX typesetting in this forum. This would make you posts, if not more useful, at least more readable

 

[The Highlander]

Instead of contributing to old threads you could spend time on learning KaTeX typesetting in this forum. This would make you posts, if not more useful, at least more readable

I already tried to persuade @scottbrandon to use LaTex in his posts (in This Post, qv) but my advice was completely ignored!

He seems determined to dig out as many old posts as he can find to add his tuppenceworth (often with no useful addition to what's already in the, sometimes ancient, threads he revives) which makes me wonder if he's just attempting to raise his "ranking" in the forum (an à la logistic_guy tactic?).

Maybe he will consider taking your advice this time?

 

[mrtwhs]

Factoring out n/4,
[math]T =\frac{n}{4}\Big(4a^3+ 6a^2d(n-1)+ 2ad^2(n-1)(2n-1)+ d^3n(n-1)^2\Big)[/math]
At this stage, I tried to factor out the known sum
[math]S = \frac{n}{2}\big(2a+(n-1)d\big)[/math]

Rewrite
[math]T =\frac{n}{2}\Big(2a^3+ 3d(n-1)a^2+ d^2(n-1)(2n-1)a+ \frac{1}{2}d^3n(n-1)^2\Big)[/math]Now divide longhand by [imath]S[/imath] thinking of [imath]T[/imath] as a polynomial in [imath]a[/imath]. The [imath]\frac{n}{2}[/imath] cancels and the rest comes out evenly! Be sure to use lots of paper and a pencil with a good eraser as well! I just tried it and it appears to work.

 

[Aion]

Rewrite
[math]T =\frac{n}{2}\Big(2a^3+ 3d(n-1)a^2+ d^2(n-1)(2n-1)a+ \frac{1}{2}d^3n(n-1)^2\Big)[/math]Now divide longhand by [imath]S[/imath] thinking of [imath]T[/imath] as a polynomial in [imath]a[/imath]. The [imath]\frac{n}{2}[/imath] cancels and the rest comes out evenly! Be sure to use lots of paper and a pencil with a good eraser as well! I just tried it and it appears to work.

I was thinking along similar lines when I reread the problem. We can also use the factor theorem to show that the division has zero remainder.


Treating [imath]n[/imath] and [imath]d[/imath] as fixed, regard [imath]S[/imath] and [imath]T[/imath] as polynomials in [imath]a[/imath]. Write

[math]S(a)=\frac{n}{2}\bigl(2a+(n-1)d\bigr)[/math]
and

[math]T(a)=\frac{n}{2}p(a)[/math]
where

[math]p(a)=2a^3+3d(n-1)a^2+d^2(n-1)(2n-1)a+\frac{1}{2}d^3n(n-1)^2[/math]
If [imath]S(a)[/imath] is indeed a factor of [imath]T(a)[/imath], then every root of [imath]S(a)[/imath] should also be a root of [imath]T(a)[/imath]. Setting [imath]S(a)=0[/imath] gives

[math]2a+(n-1)d=0[/math]
so let

[math]r=-\frac{(n-1)d}{2}[/math]
Substituting [imath]r[/imath] into [imath]p[/imath] gives

[math]p(r)=0[/math]
Therefore, by the factor theorem

[math]p(a)=(a-r)q(a)[/math]
for some polynomial [imath]q(a)[/imath]. Hence

[math]T(a)=\frac{n}{2}(a-r)q(a)[/math]
Also

[math]a-r=a+\frac{(n-1)d}{2} =\frac{2a+(n-1)d}{2}[/math]
So

[math]S(a)=\frac{n}{2}\bigl(2a+(n-1)d\bigr) =n(a-r)[/math]
Consequently

[math]T(a)=\frac{n}{2}(a-r)q(a) =S(a)\frac{q(a)}{2}[/math]
Thus [imath]S(a)[/imath] divides [imath]T(a)[/imath].

I also tried the long division as you suggested, and it indeed comes out evenly. Having worked through both methods, I actually found the long division quicker than calculating [imath]p(r)[/imath] and simplifying it all the way to [imath]0[/imath]. The factor-theorem approach is nevertheless a nice way to explain why the division has zero remainder. In retrospect, Nasi112's suggestion to calculate [imath]T/S[/imath] was actually a good idea. The mistake was in treating a quotient of two sums as a sum of quotients; carrying out the polynomial division instead gives the correct quotient.