# How can I prove the sum of squares and the sum of cubes with binomial coefficients? [closed]

My main problem is starting. I can't "see" anything that might give me an idea to find a relationship between these two things

Thank you :)

## closed as unclear what you're asking by JonMark Perry, iadvd, астон вілла олоф мэллбэрг, C. Falcon, user91500Feb 8 '17 at 8:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• What makes you think this must be proved with binomial coefficients? You asked a very similar question here last hour, which pretty much settled that matter. – dxiv Feb 6 '17 at 6:52
• – lab bhattacharjee Feb 6 '17 at 6:53
• I'm also unsure of the relationship between binomial coefficients and these sums, but it is the case that $\sum_{k=1}^n k = \frac{n(n+1)}{2} = \binom{n+1}{2}$ and $\sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \binom{n+1}{2}^2$, so maybe that can be a starting point? In what context did you encounter this problem? – CodeLabMaster Feb 6 '17 at 6:54
• Possible duplicate of Sum of cubes of binomial coefficients – GNUSupporter 8964民主女神 地下教會 Feb 6 '17 at 7:12
• @GNUSupporter no, that question is different. – The Bosco Feb 6 '17 at 7:18

If I may hazard a guess, perhaps you are expected to use the identities $$\sum_{k=0}^n\binom k2=\binom{n+1}3$$ and $$\sum_{k=0}^n\binom k3=\binom{n+1}4$$ to derive formulas for $$\sum_{k=0}^nk^2$$ and $$\sum_{k=0}^nk^3.$$ For instance, $$\binom{n+1}3=\sum_{k=0}^n\binom k2=\sum_{k=0}^n\frac{k^2-k}2=\frac12\sum_{k=0}^nk^2-\frac12\sum_{k=0}^nk$$ so $$\sum_{k=0}^nk^2=2\binom{n+1}3+\sum_{k=0}^nk=2\binom{n+1}3+\binom{n+1}2.$$

The binomial coefficients appear in the expansion of powers, such as $(k-1)^2=k^2-2k+1$.

Now consider

$$\sum_{k=1}^nk^2-\sum_{k=1}^n(k-1)^2.$$

One one hand, this difference is the single term $n^2$. On the other, is is a linear combination of sums of $k^d$ for $0\le d<2$ (the terms $k^2$ cancel out):

$$n^2=2\sum_{k=1}^nk-\sum_{k=1}^n1,$$ from wich you draw

$$S_2(n)=\sum_{k=1}^nk^2=\frac{n^2+n}2.$$

You can repeat the reasoning with

$$n^3=3\sum_{k=1}^nk^2-3\sum_{k=1}^nk+\sum_{k=1}^n1$$ and $$n^4=4\sum_{k=1}^nk^3-6\sum_{k=1}^nk^2+4\sum_{k=1}^nk-\sum_{k=1}^n1.$$

To prove $\sum_{k=1}^n k^3 = \binom{n+1}{2}^2$, use Pascal's law to find

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

and use induction. You're likely to find the formula for sums of squares in a similar fashion.