How to prove that $\sum_{M=k-1}^{2 k-1}\left(q C_{M}^{k-1} q^{k-1} p^{M-(k-1)}+p C_{M}^{k} p^{k} q^{M-k}\right)=1$ where $q+p = 1$

Question

How to prove $\sum_{M=k-1}^{2 k-1}\left(q C_{M}^{k-1} q^{k-1} p^{M-(k-1)}+p C_{M}^{k} p^{k} q^{M-k}\right)=1$ ,where $q+p = 1$. And here $C_{m}^{n}=\frac{m !}{n !(m-n) !} $ for $m \geq n$, otherwise $0$. I've tried to expand the sum and considered to use the Taylor series in two variables. But I couldn't get the right answer. Am I thinking in a wrong way?

Answer 1

Here we seek to prove that

$$\sum_{M=k-1}^{2k-1} \left({M\choose k-1} (1-x)^k x^{M-(k-1)} + {M\choose k} x^{k+1} (1-x)^{M-k}\right) = 1.$$

The LHS is

$$\sum_{q=0}^n {q+n-1\choose n-1} (1-x)^n x^q + \sum_{q=1}^n {q+n-1\choose n} x^{n+1} (1-x)^{q-1} \\ = (1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q + x^{n+1} \sum_{q=1}^n {q+n-1\choose n} (1-x)^{q-1}.$$

We have for the first sum

$$ [z^n] \frac{1}{1-z} \sum_{q\ge 0} {q+n-1\choose n-1} x^q z^q = [z^n] \frac{1}{1-z} \frac{1}{(1-xz)^n} \\ = \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{1-z} \frac{1}{(1-xz)^n}.$$

Residues sum to zero and fortunately the residue at infinity is zero by inspection. Therefore we require the residues at one and at $z=1/x.$ We start with

$$(-1)^{n+1} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{z-1} \frac{1}{(xz-1)^n} \\ = \frac{(-1)^{n+1}}{x^n} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}} \frac{1}{z-1} \frac{1}{(z-1/x)^n}.$$

The residue at one is

$$\frac{(-1)^{n+1}}{x^n} \frac{1}{(1-1/x)^n} = \frac{(-1)^{n+1}}{(x-1)^n}.$$

Multiply by the factor in front to get a contribution of $-1.$ For the residue at $z=1/x$ we require

$$\frac{1}{(n-1)!} \left(\frac{1}{z^{n+1}} \frac{1}{z-1}\right)^{(n-1)} \\ = \frac{1}{(n-1)!} \sum_{q=0}^{n-1} {n-1\choose q} \frac{(-1)^q (n+q)!}{n! z^{n+1+q}} \frac{(-1)^{n-1-q} (n-1-q)!}{(z-1)^{n-q}} \\ = (-1)^{n-1} \sum_{q=0}^{n-1} {n+q\choose n} \frac{1}{z^{n+1+q} (z-1)^{n-q}}.$$

The residue is (evaluate at $z=1/x$)

$$\frac{(-1)^{n+1}}{x^n} (-1)^{n-1} \sum_{q=0}^{n-1} {n+q\choose n} x^{n+1+q} \frac{1}{(1/x-1)^{n-q}} \\ = \sum_{q=0}^{n-1} {n+q\choose n} x^{q+1} \frac{x^{n-q}}{(1-x)^{n-q}} \\ = x^{n+1} \sum_{q=0}^{n-1} {n+q\choose n} \frac{(1-x)^q}{(1-x)^{n}}.$$

Multiply by the factor in front and shift the index to get a contribution of

$$x^{n+1} \sum_{q=1}^{n} {n+q-1\choose n} (1-x)^{q-1}.$$

Using the fact that residues sum to zero we have shown that

$$(1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q -1 + x^{n+1} \sum_{q=1}^n {q+n-1\choose n} (1-x)^{q-1} = 0.$$

which is the claim.

Remark. This identity turns out to be an identity by Gosper which was proved at the following MSE link. To see this rewrite as

$$(1-x)^n \sum_{q=0}^n {q+n-1\choose n-1} x^q + x^{n+1} \sum_{q=0}^{n-1} {q+n\choose n} (1-x)^{q} = 1.$$

The original is

$$\sum_{q=0}^{m-1} {n-1+q\choose q} x^n (1-x)^q + \sum_{q=0}^{n-1} {m-1+q\choose q} x^q (1-x)^m = 1.$$

Now replace $m$ by $n$ and $n$ by $n+1.$

Answer 2

FYI, have a look at the formation:

Consider a game: there are two chessman $A$ and $B$ racing along the axis. $A$ starts at coordinate $0$ and $B$ starts at coordinate $1$. Every time you pull the trigger, there is possibility $p$ that $A$ moves $1$ step rightwards (while $B$ stays still), otherwise (with possibility $q=1-p$) $B$ moves $1$ step rightwards (while $A$ stays still). Once a chessman reaches coordinate $k+1$, it wins and game's over.

Suppose you pulled the trigger for $M$ times before the final straw. Apparently $M\in [k-1,2k-1]$: $k-1$ for $A=0,B=k$ and $2k-1$ for $A=k,B=k$. Now $\forall M$ there are two cases to terminate the game:

• A wins. $i.e.$ $M$ right-steps are assigned $s.t.$ $A=k,B=1+M-k$ and the final step goes to $A$. The probability of this case is $pC_M^kp^kq^{M-k}$.

• B wins. $i.e.$ $M$ right-steps are assigned $s.t.$ $B=k,A=M-(k-1)$ and the final step goes to $B$. The probability of this case is $qC_M^{k-1}q^{k-1}p^{M-k+1}$.

With $M$ traversing among range $[k-1,2k-1]$ and both the two cases, all the possibilities are enumerated completely. Hence LHS$=1$.