# Closed form expression for unusual sum of binomial coefficients

How do I get a closed form expression for $\sum_{i=c}^{n} i\binom{i}{c}$? Note that the index ranges over the upper values of the binomial, not the lower.

I know computer algebra systems can give me an answer and I can then verify it using induction, but that's not what I want: I want to derive the closed form expression without using knowledge about what the answer looks like.

• The question is not so clear.. Do you really mean $\sum _{i=c}^{n} i \binom {i}{c}$? – user87543 Dec 4 '13 at 7:01
• Yes, I really mean that. – John Dec 4 '13 at 7:10
• The way computer algebra systems derive the closed form expression is by using knowledge about what the answer looks like. Specifically, if your expression has an indefinite sum then it's the term multiplied by a rational polynomial, and it's possible to bound the degrees of the numerator and denominator. See Gosper's algorithm – Peter Taylor Dec 4 '13 at 12:19
• I've read A=B, and it's a fascinating book. I was however hoping for a more human friendly proof here. – John Dec 4 '13 at 12:42

Note that the summation can be taken to start at $i=0$ with no change, as all added terms are$~0$.

After some playing around with this summation, I found the following approach the most useful. Without the factor $i$, one knows how to do partial sums of columns in Pascal's triangle $$\sum_{i=0}^n\binom ic=\binom{n+1}{c+1},$$ and more generally, taking differences, $$\sum_{i=k+1}^n\binom ic=\binom{n+1}{c+1}-\binom{k+1}{c+1}.$$ Now realising that $i=\#\{\,k\in\Bbb N\mid k<i\,\}$, we can reorganise our summation \begin{align} \sum_{i=0}^ni\binom ic & =\sum_{i=0}^n\sum_{k<i}\binom ic =\sum_{k=0}^{n-1}\sum_{i=k+1}^n\binom ic \\& =\sum_{k=0}^{n-1}(\binom{n+1}{c+1}-\binom{k+1}{c+1}) \\& =n\binom{n+1}{c+1}-\sum_{k=0}^{n-1}\binom{k+1}{c+1} \\& =n\binom{n+1}{c+1}-\binom{n+1}{c+2}. \end{align}

Added. Thinking about it I realise there is better than this ad hoc approach. The main difficulty with the question as formulated is that the factor $i$ increases in the wrong direction so that it is not a convolution. The recipe that will treat any summing any polynomial of the upper index times a column of Pascal's triangle is

1. Write the polynomial factor in terms for the distance to the final index, so as to get a convolution $\sum_iP(i)\binom{n-i}c$;
2. Express the polynomial $P(i)$ as combination of binomial coefficients $\binom ik$;
3. For each term, apply the upper-index variation of the Vandermonde convolution: $$\sum_{i=0}^n\binom ik\binom{n-i}c=\binom{n+1}{k+c+1}.$$

In the example at hand, step 1. gives $\sum_{i=0}^n(n-i)\binom{n-i}c$, step 2. gives $n-i=n\binom i0-\binom i1$, and step 3. gives \begin{align} \sum_{i=0}^n(n-i)\binom{n-i}c& =n\sum_{i=0}^n\binom i0\binom{n-i}c-\sum_{i=0}^n\binom i1\binom{n-i}c \\& =n\binom{n+1}{c+1}-\binom{n+1}{c+2}. \end{align}

• Thanks, this is exactly what I was looking for. There's something confusing however: the rule is usually stated $\sum_{i=c}^{n}\binom{i}{c} = \binom{n+1}{c+1}$, that is, indexing from i=c up to n (rather than i=0..n). I'm not sure if this is an omission, or that one should interpret binomials as zero whenever i < c, but it seems unnecessarily confusing, especially considering the otherwise very elegant derivation. – John Dec 4 '13 at 13:23
• My first sentence (recently added) says you can make the sum start at $0$ with no change. Indeed $\binom ic=0$ when $0\leq i<c$ is a fact, not just a matter of interpretation. I usually think of $\binom{n+1}{k+1}$ as giving the sum of the entire column $k$, from $\binom0k$ down to $\binom nk$, just like the finite differences of column $k+1$ give the binomial coefficients in column $k$ everywhere, not just in the nonzero part of the triangle. But this is largely a matter of taste. – Marc van Leeuwen Dec 4 '13 at 13:32
• Thanks for clarification and an even better proof. I've accepted this answer. – John Dec 4 '13 at 14:51

$$\small \begin{matrix} . & . & . & . & . & . & | & 1 & . & . & . & . & . & | \\ . & . & . & . & . & . & | & 1 & 1 & . & . & . & . & | \\ . & . & . & . & . & . & | & 1 & 2 & 1 & . & . & . & | \\ . & . & . & . & . & . & | & 1 & 3 & 3 & 1 & . & . & | \\ . & . & . & . & . & . & | & 1 & 4 & 6 & 4 & 1 & . & | \\ . & . & . & . & . & . & | & 1 & 5 & 10 & 10 & 5 & 1 & | \\ - & - & - & - & - & - & + & - & - & - & - & - & - & + \\ . & 1 & 2 & 3 & 4 & 5 & | & y_0 & y_1 & y_2 & y_3 & y_4 & y_5 & | \\ - & - & - & - & - & - & + & - & - & - & - & - & - & + \end{matrix}$$ where the number of columns of the left vector (and the number of rows of the matrix) is given in the value of the indeterminate $n$... (If the OP wants, he can insert that scheme in his question...)
ALso, there was an answer given which points to the derivatives of some function, which I think is a good hint: the left vector could be written as $[0,1x,2x^2,3x^3,...,nx^n]$ and then one could observe the occuring expressions for the sum (getting $y_0$ using column 0 of the matrix), find a general expression for them and their derivatives (getting $y_k \cdot k!$ using the other columns of the matrix), and then one would insert $1$ for $x$...(but I think it's even easier when detecting the type of sum which is involved here because it is a very usual one...)
A further hint: note that the dotproduct $y_0$ seen as function of $x$, where $x$ is inserted in the left vector as indicated above, gives the expression: $$y_0(x,n) =x {1-x^n\over (1-x)^2} - nx \cdot{ x^n \over (1-x)}$$ (which has be derived from the expression for the finite geometric sum and their derivative).
To evaluate it (and the derivatives $$y_1(x,n)=y_0(x,n)'/1! \\ y_2(x,n)=y_0(x,n)''/2! \\ ...$$) at $x=1$ one has to cancel the $(1-x)$-factor and then, I think, one has to use L'Hospital's rule - but I didn't proceed to this for the moment)