# A general formula for $\sum (k-1)(k-2)(k-3)$?

What is a "simpler" formula for

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

• Expand the polynomial, separate out the terms, and use the formulae for sums of powers. Commented Dec 22, 2010 at 5:22
• The answer to the question asked is "Yes" Commented Dec 22, 2010 at 5:36
• @Ross: My mistake. Let me try to correct the question statement. Commented Dec 22, 2010 at 6:01
• 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. Commented Dec 22, 2010 at 6:47
• ...or read Section 2.6 in Concrete Mathematics by Graham, Knuth & Patashnik. Commented Dec 22, 2010 at 7:38

Hint:

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

• 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.) Commented Dec 22, 2010 at 6:18
• $\frac{k(k-1)(k-2)(k-3)}{4}$, is this right?
– qed
Commented Dec 24, 2010 at 5:40
• @Craving: You mean $n$ instead of $k$, right? Also, you are forgetting the $6$ in the denominator. Commented Dec 24, 2010 at 7:09

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...?

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

• 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}]] ? Commented Dec 22, 2010 at 6:04
• @Debanjan: You're missing $\frac{1}{6}$. Also, your first sum should be up to $n-2$, not $n$. Commented Dec 22, 2010 at 6:04
• Opps sure $\frac{1}{6}$! but why $n-2$ ? Commented Dec 22, 2010 at 6:06
• @Debanjan: Why does the first sum start at $1$ instead of $3$? Commented Dec 22, 2010 at 6:08
• 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? Commented Dec 22, 2010 at 6:12

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 $\sum_{m=0}^n m^k$.

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.

• 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.) Commented Dec 22, 2010 at 7:44
• @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? Commented Dec 22, 2010 at 22:02
• It has several names: see de.wikipedia.org/wiki/Fallende_Faktorielle and also en.wikipedia.org/wiki/Pochhammer_symbol. Commented Dec 22, 2010 at 22:28

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.

• Also check out this post of mine for some more information about telescoping. Commented Dec 22, 2010 at 21:23

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 $$\binom{n-1-3+5-1}{5-1}=\binom{n}{4},\quad n\geqslant3.$$