How can I prove the sum of squares and the sum of cubes with binomial coefficients?

Question

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 :)

Answer 1

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

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

Answer 3

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.