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What is a "simpler" formula for

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

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    $\begingroup$ Expand the polynomial, separate out the terms, and use the formulae for sums of powers. $\endgroup$ – J. M. is a poor mathematician Dec 22 '10 at 5:22
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    $\begingroup$ The answer to the question asked is "Yes" $\endgroup$ – Ross Millikan Dec 22 '10 at 5:36
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    $\begingroup$ @Ross: My mistake. Let me try to correct the question statement. $\endgroup$ – Aryabhata Dec 22 '10 at 6:01
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    $\begingroup$ If you are more interested, there is a tool called finite calculus and this is an example of a sum of a falling power. See this document: stanford.edu/~dgleich/publications/finite-calculus.pdf. $\endgroup$ – J. J. Dec 22 '10 at 6:47
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    $\begingroup$ ...or read Section 2.6 in Concrete Mathematics by Graham, Knuth & Patashnik. $\endgroup$ – Hans Lundmark Dec 22 '10 at 7:38
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Hint:

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

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    $\begingroup$ 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.) $\endgroup$ – Dave Radcliffe Dec 22 '10 at 6:18
  • $\begingroup$ $\frac{k(k-1)(k-2)(k-3)}{4}$, is this right? $\endgroup$ – qed Dec 24 '10 at 5:40
  • $\begingroup$ @Craving: You mean $n$ instead of $k$, right? Also, you are forgetting the $6$ in the denominator. $\endgroup$ – Aryabhata Dec 24 '10 at 7:09
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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...?

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

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  • $\begingroup$ Why different answer ? What I am missing : Expand[Sum[j^3 - j, {j, 1, n}]] and Expand[Sum[(k - 1) (k - 2) (k - 3), {k, 3, n}]] ? $\endgroup$ – Quixotic Dec 22 '10 at 6:04
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    $\begingroup$ @Debanjan: You're missing $\frac{1}{6}$. Also, your first sum should be up to $n-2$, not $n$. $\endgroup$ – Jonas Meyer Dec 22 '10 at 6:04
  • $\begingroup$ Opps sure $\frac{1}{6}$! but why $n-2$ ? $\endgroup$ – Quixotic Dec 22 '10 at 6:06
  • $\begingroup$ @Debanjan: Why does the first sum start at $1$ instead of $3$? $\endgroup$ – Jonas Meyer Dec 22 '10 at 6:08
  • $\begingroup$ I really don't understand what you mean... for the first we have to compute the sum of all values of k from $3$ to $n$ right? $\endgroup$ – Quixotic Dec 22 '10 at 6:12
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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 $\sum_{m=0}^n m^k$.

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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.

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    $\begingroup$ 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.) $\endgroup$ – Hans Lundmark Dec 22 '10 at 7:44
  • $\begingroup$ @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? $\endgroup$ – Arjang Dec 22 '10 at 22:02
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    $\begingroup$ It has several names: see de.wikipedia.org/wiki/Fallende_Faktorielle and also en.wikipedia.org/wiki/Pochhammer_symbol. $\endgroup$ – Hans Lundmark Dec 22 '10 at 22:28
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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.

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    $\begingroup$ Also check out this post of mine for some more information about telescoping. $\endgroup$ – Matt Calhoun Dec 22 '10 at 21:23
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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 $$\binom{n-1-3+5-1}{5-1}=\binom{n}{4},\quad n\geqslant3.$$

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