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How to find the summation of $\Sigma\frac{(n+1)(n+2)}{2!}$ ?

MY WORK:

I know that the expression in the summation is the general term of the binomial expansion $(1-x)^{-3}$ . I have a solution where I consider : $$(1-x)^{-3}=a_0+a_1x+a_2x^2 +...+a_nx^n +...$$ Then I multiply it with the following : $$(1-x)^{-1}=1+x+x^2+x^3+ ... + x^n+ ...$$ Equating co efficients of $x^n$ , I get : $$a_0+a_1+a_2+...+a_n=\Sigma\frac{(n+1)(n+2)}{2!}=\frac{(n+1)(n+2)(n+3)}{3!}$$ ...

CAN I DO IT IN ANY OTHER WAY OTHER THAN BINOMIAL EXPANSION?

Michael Rozenberg
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Entrepreneur
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  • summation over what? In general $\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$. Can that be used here with $r=2$? – drhab Jun 26 '18 at 08:38
  • Related : https://math.stackexchange.com/questions/1490794/proof-of-the-hockey-stick-identity-sum-limits-t-0n-binom-tk-binomn1 – Arnaud D. Oct 29 '18 at 11:15

3 Answers3

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$$\sum_{k=0}^n\frac{(k+1)(k+2)}{2}=\sum_{k=0}^n\left(\frac{(k+1)(k+2)(k+3)}{6}-\frac{k(k+1)(k+2)}{6}\right)=\frac{(n+1)(n+2)(n+3)}{6}-0=\frac{(n+1)(n+2)(n+3)}{6}.$$

Michael Rozenberg
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You can do it like this:\begin{align}\sum_{k=0}^n\frac{(k+1)(k+2)}2&=\frac12\sum_{k=0}^nk^2+\frac32\sum_{k=0}^nk+\sum_{k=0}^n1\\&=\frac{n(n+1)(2n+1)}{12}+\frac{3n(n+1)}4+n+1.\end{align}Now, all it takes is to check that$$\frac{n(n+1)(2n+1)}{12}+\frac{3n(n+1)}4+n+1=\frac{(n+1)(n+2)(n+3)}6.$$

José Carlos Santos
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  • Yes, that is possible, but IMHO expressions like $\sum_{k=0}^nk^r$ must be deduced from the more friendly expressions $\sum_{k=r}^{n}\binom{k}{r}=\binom{n+1}{r+1}$ which moreover are easy to prove by induction. Not the other way around. – drhab Jun 26 '18 at 09:01
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Under the convention that for nonnegative integers $r,k$: $$\binom{k}{r}=0\text{ if }k<r$$ it can be proved by induction that: $$\sum_{k\leq n}\binom{k}{r}=\binom{n+1}{r+1}$$

Induction step:$$\sum_{k\leq n}\binom{k}{r}=\sum_{k\leq n}\binom{k-1}{r-1}+\sum_{k\leq n}\binom{k-1}{r}=\binom{n}{r}+\binom{n}{r+1}=\binom{n+1}{r+1}$$

For $r=2$ we find: $$\sum_{k=0}^{n}\frac{(k+1)(k+2)}2=\sum_{k=0}^{n}\binom{k+2}{2}=\sum_{k\leq n+2}\binom{k}{2}=\binom{n+3}{3}=\frac{(n+1)(n+2)(n+3)}6$$

drhab
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