Express the sum of the sequence of cubes as a polynomial in n using the summation on the upper index formula: $$ \sum\limits_{k=0}^n\binom{k}{m} = \binom{n+1}{m+1} $$ It has been proven that the sum of sequence of cubes can be expressed as the following fraction: $$ \sum\limits_{k=1}^n(k^3) = \frac{n^2(n+1)^2}{4} $$ However, I am stuck and can't find a way to apply the said formula to get the fraction. I guess I haven't grasped yet the connection between binomial coefficients and powers. Can I get a hint?

  • $\begingroup$ You want to prove the middle formula via the binomial sum equality? $\endgroup$
    – user144248
    Sep 22 '14 at 19:06
  • $\begingroup$ @user144248 Yes, exactly! $\endgroup$
    – Baldee
    Sep 22 '14 at 19:16
  • $\begingroup$ $k^3 = 6{k \choose 3} + 3 k^2 -2k $ (I got this by expanding out ${k \choose 3}$). Now eliminate the $k^2$ using ${k \choose 2}$. Then again for the leftover $k$ term. Now, if you sum the LHS, you sum the RHS. But by the stated thm, you can calculate the sum of the RHS. $\endgroup$
    – Mitch
    Sep 25 '14 at 15:55


With the formula we find:


This enables you to express $\sum_{k=0}^{n}k^{3}$ in formulas of $\sum_{k=0}^{n}k^{2}$ and $\sum_{k=0}^{n}k$.

You can use a version of the formula again ($m=2$) to find likewise a formula for $\sum_{k=0}^{n}k^{2}$ in $\sum_{k=0}^{n}k$.


  • $\binom{n+1}{2}=\sum_{k=0}^{n}\binom{k}{1}=\sum_{k=0}^{n}k$

  • $\binom{n+1}{3}=\sum_{k=0}^{n}\binom{k}{2}=\frac{1}{2}\sum_{k=0}^{n}k\left(k-1\right)=\frac{1}{2}\sum_{k=0}^{n}k^{2}-\frac{1}{2}\sum_{k=0}^{n}k$

  • $\binom{n+1}{4}=\sum_{k=0}^{n}\binom{k}{3}=\frac{1}{6}\sum_{k=0}^{n}k\left(k-1\right)\left(k-2\right)=\frac{1}{6}\sum_{k=0}^{n}k^{3}-\frac{3}{6}\sum_{k=0}^{n}k^{2}+\frac{2}{6}\sum_{k=0}^{n}k$

  • $\begingroup$ How do we get to $\sum\limits_{k=0}^n\binom{k}{3}$ from $ \sum\limits_{k=0}^n(k^3)$ ? $\endgroup$
    – Baldee
    Sep 22 '14 at 19:34
  • $\begingroup$ You are asked to find a formula for $\sum k^{3}$, aren't you? Then you are supposed to do the opposite: get to $\sum k^{3}$ from $\sum\binom{k}{3}$. My answer gives a hint how to do that. $\endgroup$
    – drhab
    Sep 22 '14 at 20:28

Using a difference table for $(n^3)$, we get






Then $n^3=0\binom{n}{0}+1\binom{n}{1}+6\binom{n}{2}+6\binom{n}{3}$,

so $\displaystyle\sum_{k=0}^{n}k^3=\sum_{k=0}^{n}\left[1\binom{k}{1}+6\binom{k}{2}+6\binom{k}{3}\right]=1\binom{n+1}{2}+6\binom{n+1}{3}+6\binom{n+1}{4}$

using the given summation formula.

Now show that this gives the stated formula for $\displaystyle\sum_{k=1}^nk^3$.

  • $\begingroup$ Thank you! I do understand the second equality, but I do not understand how we express $n^3$ as a sum of binomial coefficients. Can you clarify? $\endgroup$
    – Baldee
    Sep 22 '14 at 20:09
  • $\begingroup$ The easiest way I know to do this is to construct the difference table (by writing the terms of the sequence in the first line, and then continuing to take differences to get the succeeding lines), and then using the left diagonal to get the coefficients. You could also do this by setting $n^3=a\binom{n}{0}+b\binom{n}{1}+c\binom{n}{2}+d\binom{n}{3}$ and then substituting $n=0,1,2,3$ to solve for the coefficients. $\endgroup$
    – user84413
    Sep 22 '14 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.