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What is a "simpler" formula for

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

  • 0
    @user5035: Could you please elaborate on your question? What have you tried? Hint: One way to do this uses the formulas seen at http://pirate.shu.edu/~wachsmut/ira/infinity/answers/sm_sq_cb.html2010-12-22
  • 1
    Expand the polynomial, separate out the terms, and use the formulae for [sums of powers](http://www.mathpages.com/home/kmath279/kmath279.htm).2010-12-22
  • 4
    The answer to the question asked is "Yes"2010-12-22
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    @Debanjan: Your formula is not quite right.2010-12-22
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    @Ross: My mistake. Let me _try_ to correct the question statement.2010-12-22
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    Yes Jonas, I was going to ask you that.2010-12-22
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    Corrected formula : $\frac{1}{24} (n^4 - 6 n^3+ 11 n^2 -6 n)$2010-12-22
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    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: http://www.stanford.edu/~dgleich/publications/finite-calculus.pdf.2010-12-22
  • 3
    ...or read Section 2.6 in *Concrete Mathematics* by Graham, Knuth & Patashnik.2010-12-22
  • 0
    Many good answers, I really don't know which one to pick.2010-12-24

7 Answers 7

16

Hint:

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

  • 6
    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.)2010-12-22
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    $\frac{k(k-1)(k-2)(k-3)}{4}$, is this right?2010-12-24
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    @Craving: You mean $n$ instead of $k$, right? Also, you are forgetting the $6$ in the denominator.2010-12-24
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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...?

6

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

  • 0
    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}]]` ?2010-12-22
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    @Debanjan: You're missing $\frac{1}{6}$. Also, your first sum should be up to $n-2$, not $n$.2010-12-22
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    Opps sure $\frac{1}{6}$! but why $n-2$ ?2010-12-22
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    @Debanjan: Why does the first sum start at $1$ instead of $3$?2010-12-22
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    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?2010-12-22
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    @Debanjan: I mean the first sum in your comment, Sum[j^3 - j, {j, 1, n}]. You correctly have $j$ starting at $1$ instead of $3$, for the same reason you should have it ending at $n-2$ rather than $n$.2010-12-22
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    Oh sorry! now I get your point.Thanks a lot :-)2010-12-22
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.

  • 1
    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.)2010-12-22
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    @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?2010-12-22
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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.2010-12-22
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 $\sum_{m=0}^n m^k$.

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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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    Also check out [this](http://math.stackexchange.com/questions/13701/mathematical-telescoping) post of mine for some more information about telescoping.2010-12-22
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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 $$\binom{n-1-3+5-1}{5-1}=\binom{n}{4},\quad n\geqslant3.$$