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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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closed as unclear what you're asking by JonMark Perry, iadvd, астон вілла олоф мэллбэрг, C. Falcon, user91500 Feb 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.

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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.

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