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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 :)

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    What makes you think this *must* be proved with binomial coefficients? You asked a very similar question [here](http://math.stackexchange.com/questions/2131354/how-do-i-expand-a-binomial-coefficient-into-a-closed-form-function) last hour, which pretty much settled *that* matter. – 2017-02-06
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    http://math.stackexchange.com/questions/134747/sum-of-cubes-of-binomial-coefficients http://math.stackexchange.com/questions/586138/strehl-identity-for-the-sum-of-cubes-of-binomial-coefficients http://planetmath.org/sumofpowersofbinomialcoefficients – 2017-02-06
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    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? – 2017-02-06
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    Possible duplicate of [Sum of cubes of binomial coefficients](http://math.stackexchange.com/questions/134747/sum-of-cubes-of-binomial-coefficients) – 2017-02-06
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    @GNUSupporter no, that question is different. – 2017-02-06
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    @CodeLabMaster Thank you so much. This is a problem I have to do. The context is to prove it using Pascal's identity (and in general binomial coefficients) – 2017-02-06
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    @dxiv not for me. Sorry. This is an exercise, and that is why I do not ask for a complete solution. Just a hint. I have to prove it using binomial coefficients. – 2017-02-06
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    @TheBosco You can prove $$\sum_{k=1}^n k = \binom{n+1}{2}$$ with pascal's law and induction. Perhaps you can do the same for higher exponents. – 2017-02-06
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    @TheBosco Sorry, I made a mistake while reading the title. – 2017-02-07

3 Answers 3

2

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 $$\sum_{k=0}^nk^2=2\binom{n+1}3+\sum_{k=0}^nk=2\binom{n+1}3+\binom{n+1}2.$$

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

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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.