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Newton's interpolation formula shows that every polynomial is a linear combination of binomial coefficients. For instance, $$ \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} = 0 \binom{n}{0}+1\binom{n}{1}+3\binom{n}{2}+2\binom{n}{3} $$ Using Pascal's relation, this can be reduced $$ \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\binom{n+2}{3}+\binom{n+1}{3} $$ Is this the simplest linear combination of binomial coefficients for that polynomial?

Here is another example: $$ n^5-5n^3+4n = 120 \binom{n}{3} + 240 \binom{n}{4} + 120 \binom{n}{5} $$ can be reduced to $$ n^5-5n^3+4n = 120 \binom{n+2}{5} $$

Is there a systematic method to find these economical representations?

lhf
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  • Examples from https://math.stackexchange.com/a/2350033/589 and https://math.stackexchange.com/a/1851131/589. I've used this technique in several other answers here. – lhf Jul 08 '17 at 01:00
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    I do believe that the sum of n^2 is 1/4*${2n + 2}\choose{3}$ – Vasting Jul 08 '17 at 01:04
  • @Vasting, oh, right! Which is my point: how do we simplify linear combination of binomial coefficients? Please add your comment as an answer. It does answer half of my question. – lhf Jul 08 '17 at 01:05

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