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.
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.