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What is a "simpler" formula for

$$\sum_{3}^{n} \frac{(k-1)(k-2)(k-3)}{6}$$

J. M. ain't a mathematician
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qed
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7 Answers7

16

Hint:

$$k(k-1)(k-2)(k-3) - (k-1)(k-2)(k-3)(k-4) = 4(k-1)(k-2)(k-3)$$

Aryabhata
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    Let $f(k) = k(k-1)(k-2)(k-3)$. Then $f(k) - f(k-1) = 4(k-1)(k-2)(k-3)$. Summing both sides from $k=3$ to $n$ gives the result. This is an example of a telescoping series. (Do a web search for this term if it is unfamiliar.) – Dave Radcliffe Dec 22 '10 at 06:18
  • $\frac{k(k-1)(k-2)(k-3)}{4}$, is this right? – qed Dec 24 '10 at 05:40
  • @Craving: You mean $n$ instead of $k$, right? Also, you are forgetting the $6$ in the denominator. – Aryabhata Dec 24 '10 at 07:09
11

Show that ${k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_4 = k$ in non-negative integers. Then $\sum_{k=0}^{n-4} {k+3 \choose 3}$ is the number of solutions to $x_1 + ... + x_5 = n-4$ in non-negative integers, which is...?

Qiaochu Yuan
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6

I really like Aryabhata's hint, but another way to simplify the sum is to reindex with $j=k-2$, and use $(j+1)j(j-1)=j^3-j$.

J. M. ain't a mathematician
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Jonas Meyer
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4

There is no need for a general formula for this special case, all one needs is to use the formulas for $\sum {k} , \sum{k^2} \text{ and } \sum{k^3}$. Just expand the expression and use the simpler known sums.

Yuval Filmus
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jimjim
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    I would say that it's the sum in the question that is the simpler one, since it's a "falling power", and falling powers behave nicer than ordinary powers w.r.t. taking sums and differences (just like ordinary powers are nice when it comes to integrals and derivatives). (See Moron's answer.) – Hans Lundmark Dec 22 '10 at 07:44
  • @Hans, By "falling power", do you mean descending powers? Also if the original term you used for "falling power" is in German could please say the German phrase? – jimjim Dec 22 '10 at 22:02
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    It has several names: see http://de.wikipedia.org/wiki/Fallende_Faktorielle and also http://en.wikipedia.org/wiki/Pochhammer_symbol. – Hans Lundmark Dec 22 '10 at 22:28
4

Consider the following sum $$\sum_{m=0}^n \binom{m}{k}.$$ It counts the number of possibilities to select $k$ elements from at most $n$ elements. Some choices are counted multiple times, for example $\{0,\ldots,k-1\}$ is counted once for each $m \in [k-1,n]$. So it's natural to distinguish among those by tagging them with $m$ somehow. The best way to do that is to add the element $m+1$. The result is a choice of $k+1$ elements from $n+1$, and so $$\sum_{m=0}^n \binom{m}{k} = \binom{n+1}{k+1}.$$ From this formula one can extract (using linear algebra) the usual formulas for $\sum_{m=0}^n m^k$.

Yuval Filmus
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3

HINT $\ $ The sum telescopes since the falling factorial summand is a perfect difference:

$\rm\quad (k+1)^{[n]} - k^{[n]}\ =\ (k+1)\ k\ \cdots\ (k-n+2)\ -\ k\ (k-1)\ \cdots\ (k-n+1)$

$\rm\quad\phantom{(k+1)^{[n]} - k^{[n]}\ } =\ (k+1 - (k-n+1))\ \ \ k\ (k-1)\ \cdots\ (k-n+2)$

$\rm\quad\phantom{(k+1)^{[n]} - k^{[n]}\ }\ =\ n\ k^{[n-1]}$

For other examples of additive/multiplicative telescopy see here and here or here or here or here. For much more on the falling factorials see Steven Roman's textbook The Umbral Calculus.

Community
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Bill Dubuque
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2

Perhaps a messy and boring way, we can use the generating function. $$\sum_{k=3}^{n}\frac{(k-1)(k-2)(k-3)}{6}=\frac{1}{6}\sum_{k=2}^{n-1}k(k-1)(k-2)$$ In addition, the generating function for $k(k-1)(k-2)$ is $x^3\left(\frac{1}{1-x}\right)^{(3)}.$

Hence, the sum is the coefficient of $x^{n-1}$ in $\frac{1}{6}\frac{1}{1-x}x^3\left(\frac{1}{1-x}\right)^{(3)}=\frac{x^3}{(1-x)^5}$, which is $$\binom{n-1-3+5-1}{5-1}=\binom{n}{4},\quad n\geqslant3.$$

Vladimir
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