The page on Binomial Sums in Wolfram Mathworld http://mathworld.wolfram.com/BinomialSums.html (Equation 69) gives this neat-looking identity due to Gosper (1972):
$$\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]=1 $$
Would anyone know if there is a simple proof of this identity without using induction?
On
Consider the expression \begin{align} \sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k] \end{align} in the following way. First consider the generating function of the series \begin{align} S_{n} = \sum_{k=0}^{n} \binom{k+n}{k} \, x^{k} \end{align} for which it is seen that \begin{align} \phi(x, t) &= \sum_{n=0}^{\infty} S_{n} \, t^{n} \\ &= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n+k}{k} x^{k} t^{n} \\ &= \sum_{n,k=0}^{\infty} \binom{2k+n}{k} (xt)^{k} t^{n} \\ &= (1-4xt)^{-1/2} \sum_{n=0}^{\infty} \left( \frac{2t}{1+\sqrt{1-4xt}} \right)^{n} \\ &= \frac{1}{\sqrt{1-4xt}} \cdot \frac{1+\sqrt{1-4xt}}{1-2t + \sqrt{1-4xt}}, \end{align} where the series \begin{align} \sum_{k=0}^{\infty} \binom{2k+n}{k} x^{k} = \frac{2^{n}}{\sqrt{1-4x} \, (1+\sqrt{1-4x})^{n}} \end{align} was used. This is equation (66) of the site referenced in the proposed problem. Now, for $A = 1-4x(1-x)t$, \begin{align} \theta(x,t) &= x \phi(1-x, xt) + (1-x) \phi(x, (1-x)t) \\ &= \frac{1+\sqrt{A}}{\sqrt{A}} \left[ \frac{x}{1-2xt+ \sqrt{A}} + \frac{1-x}{1-2(1-x)t + \sqrt{A}} \right]. \end{align} Making use of the definition of $A$ then \begin{align} 1-2xt+\sqrt{A} = \frac{1+\sqrt{A}}{2(1-x)} \, (1-2x + \sqrt{A}) \end{align} and leads to \begin{align} \theta(x,t) &= \frac{1+\sqrt{A}}{\sqrt{A}} \left[ \frac{2x(1-x)}{(1+\sqrt{A})(1-2x+\sqrt{A}) } + \frac{2x(1-x)}{(1+\sqrt{A})(1-2(1-x)+\sqrt{A})} \right] \\ &= \frac{2x(1-x)}{\sqrt{A}} \left[ \frac{1}{1-2x+\sqrt{A}} + \frac{1}{1-2(1-x)+\sqrt{A}} \right] \\ &= \frac{4x(1-x)}{-1+4x(1-x) + A} \\ &= \left(\frac{1-A}{t}\right) \, \frac{1}{\left(\frac{1-A}{t}\right) - (1-A)} \\ &= \frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}. \end{align} Since \begin{align} \theta(x,t) &= x \phi(1-x, xt) + (1-x) \phi(x, (1-x)t) \\ &= \sum_{n=0}^{\infty} \left[ \sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k] \, \right] \, t^{n}\\ &= \sum_{n=0}^{\infty} t^{n} \end{align} then by comparison of the coefficients of $\theta(x,t)$ the result follows as \begin{align} \sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k] = 1. \end{align}
On
Let $m>0$ a natural number and let's consider a set of polynomial functions: $$A_{0}\left(x\right)=1$$ $$A_{n}\left(x\right)=\sum_{k=0}^{n}\binom{n+m}{k}x^{k}\left(1-x\right)^{n-k},\,\,\,\,n=1,\,2,\,3,\,\ldots$$ We have $$A_{l}\left(x\right)=\sum_{k=0}^{l}\binom{l-1+m}{k}x^{k}\left(1-x\right)^{l-k}+\sum_{k=1}^{l}\binom{l-1+m}{k-1}x^{k}\left(1-x\right)^{l-k}$$ $$A_{l}\left(x\right)=\left(1-x\right)A_{l-1}\left(x\right)+\binom{l-1+m}{l}x^{l}+xA_{l-1}\left(x\right)$$ $$A_{l}\left(x\right)=A_{l-1}\left(x\right)+\binom{l-1+m}{l}x^{l},\,\,\,l=1,\,2,\,\ldots,\,n$$ Then $$\sum_{l=1}^{n}A_{l}\left(x\right)=\sum_{l=0}^{n-1}A_{l}\left(x\right)+\sum_{l=1}^{n}\binom{l-1+m}{l}x^{l}$$ which leads to $$A_{n}\left(x\right)=\sum_{k=0}^{n}\binom{m+k-1}{k}x^{k}.$$ i.e. $$\sum_{k=0}^{n}\binom{n+m}{k}x^{k}\left(1-x\right)^{n-k}=\sum_{k=0}^{n}\binom{m+k-1}{k}x^{k}\, \, \left(1\right)$$ If we set $m=n+1$, we get: $$\sum_{k=0}^{n}\binom{2n+1}{k}x^{k}\left(1-x\right)^{n-k}=\sum_{k=0}^{n}\binom{n+k}{k}x^{k}$$ for any natural number $n>0$. Now $$1=\left[x+\left(1-x\right)\right]^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}x^{k}\left(1-x\right)^{2n+1-k}$$ $$\sum_{k=0}^{n}\binom{2n+1}{k}x^{k}\left(1-x\right)^{2n+1-k}+\sum_{k=n+1}^{2n+1}\binom{2n+1}{k}x^{k}\left(1-x\right)^{2n+1-k}=1$$ $$\sum_{k=0}^{n}\binom{2n+1}{k}x^{k}\left(1-x\right)^{2n+1-k}+\sum_{k=n+1}^{2n+1}\binom{2n+1}{2n+1-k}x^{k}\left(1-x\right)^{2n+1-k}=1$$ $$\sum_{k=0}^{n}\binom{2n+1}{k}x^{k}\left(1-x\right)^{2n+1-k}+\sum_{k=0}^{n}\binom{2n+1}{k}x^{2n+1-k}\left(1-x\right)^{k}=1$$ $$\left(1-x\right)^{n+1}\sum_{k=0}^{n}\binom{2n+1}{k}x^{k}\left(1-x\right)^{n-k}+x^{n+1}\sum_{k=0}^{n}\binom{2n+1}{k}\left(1-x\right)^{k}x^{n-k}=1$$ Using $\left(1\right)$ we obtain: $$\left(1-x\right)^{n+1}\sum_{k=0}^{n}\binom{n+k}{k}x^{k}+x^{n+1}\sum_{k=0}^{n}\binom{n+k}{k}\left(1-x\right)^{k}=1$$ or $$\sum_{k=0}^{n}\binom{n+k}{k}\left[x^{n+1}\left(1-x\right)^{k}+\left(1-x\right)^{n+1}x^{k}\right]=1$$ which concludes the proof.
On
Suppose we seek to show that $$\sum_{k=0}^n {n+k\choose k}(x^{n+1}(1-x)^k+(1-x)^{n+1}x^k) =1.$$
We use an Iverson bracket to control the range of $k$ so we can let it range from zero to infinity, which is
$$[[0 \le k\le n]] = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1+v+\cdots+v^n}{v^{k+1}} \; dv \\ = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{v^{n+1}-1}{(v-1)v^{k+1}} \; dv.$$
We evaluate this using the formula for the residue at infinity $$\mathrm{Res}_{z=\infty} h(z) = \mathrm{Res}_{z=0} \left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$
which in this case yields (omit the minus sign as the residues sum to zero) $$\mathrm{Res}_{v=0} \frac{1}{v^2} \frac{1/v^{n+1}-1}{(1/v-1) \times 1/v^{k+1}} = \mathrm{Res}_{v=0} \frac{1/v^{n+1}-1}{(1-v) \times 1/v^{k}} \\ = \mathrm{Res}_{v=0} \left(\frac{v^{k}}{v^{n+1}}\frac{1}{1-v} -\frac{v^k}{1-v}\right).$$
With the additional assumption that $k \ge 0$ which is the case here this yields $$\mathrm{Res}_{v=0} \frac{v^{k}}{v^{n+1}}\frac{1}{1-v} = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{v^{k}}{v^{n+1}}\frac{1}{1-v} \; dv$$ which could have been obtained by inspection.
This yields for the second component of the sum $$\frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \sum_{k\ge 0} {n+k\choose n} (1-x)^{n+1} x^{k} v^k \; dv \\ = (1-x)^{n+1} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \frac{1}{(1-xv)^{n+1}} \; dv.$$
We will evaluate this not by evaluating the residue at zero but the sum of the negatives of the residues at $v=1$ and $v=1/x$ given that the residues sum to zero.
For the residue at $v=1$ re-write the integral as follows: $$-(1-x)^{n+1} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{v-1} \frac{1}{(1-xv)^{n+1}} \; dv.$$
The residue at $v=1$ here is $$-(1-x)^{n+1} \frac{1}{(1-x)^{n+1}} = -1$$ for a contribution of $$1$$ upon negation.
For the residue at $v=1/x$ re-write the integral as follows: $$\frac{(1-x)^{n+1}}{x^{n+1}} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \frac{1}{(1/x-v)^{n+1}} \; dv \\ = (-1)^{n+1} \frac{(1-x)^{n+1}}{x^{n+1}} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \frac{1}{(v-1/x)^{n+1}} \; dv.$$
Use Leibniz' rule to differentiate the two terms in $v$ to get
$$\frac{1}{n!}\left( \frac{1}{v^{n+1}}\frac{1}{1-v} \right)^{(n)} = \frac{1}{n!} \sum_{k=0}^n {n\choose k} \frac{(-1)^k (n+k)!}{n! \times v^{n+1+k}} \frac{(n-k)!}{(1-v)^{n-k+1}}.$$
Evaluate this at $v=1/x$ including the factor in front to get for the residue $$(-1)^{n+1} \frac{(1-x)^{n+1}}{x^{n+1}} \frac{1}{n!} \sum_{k=0}^n {n\choose k} \frac{(-1)^k (n+k)!}{n! \times (1/x)^{n+1+k}} \frac{(n-k)!}{(1-1/x)^{n-k+1}} \\ = (-1)^{n+1} \frac{(1-x)^{n+1}}{x^{n+1}} \sum_{k=0}^n \frac{(-1)^k (n+k)!}{n! \times k! \times (1/x)^{n+1+k}} \frac{1}{(1-1/x)^{n-k+1}} \\ = (-1)^{n+1} (1-x)^{n+1} \sum_{k=0}^n {n+k\choose k} \frac{(-1)^k}{(1/x)^{k}} \frac{1}{(x-1)^{n-k+1}/x^{n-k+1}} \\ = \sum_{k=0}^n {n+k\choose k} \frac{(-1)^k}{(1/x)^{k}} \frac{1}{(x-1)^{-k}/x^{n-k+1}} \\ = x^{n+1} \sum_{k=0}^n {n+k\choose k} (1-x)^k.$$
Upon negation this becomes the negative of the first component of the sum. Hence adding the three pieces (first component, one, negative of first component) we obtain a sum of
$$1.$$
Remark. If we want to do this properly we also need to verify that the residue at infinity of the integral in $v$ is zero.
In the present case this becomes
$$- \mathrm{Res}_{v=0} \frac{1}{v^2} \frac{1}{(1/v)^{n+1}}\frac{1}{1-1/v} \frac{1}{(1-x/v)^{n+1}} \\ = - \mathrm{Res}_{v=0} \frac{1}{v^2} \frac{v^{n+1} \times v^{n+1}}{1-1/v} \frac{1}{(v-x)^{n+1}} \\ = - \mathrm{Res}_{v=0} \frac{1}{v} \frac{v^{2n+2}}{v-1} \frac{1}{(v-x)^{n+1}} \\ = - \mathrm{Res}_{v=0} \frac{v^{2n+1}}{v-1} \frac{1}{(v-x)^{n+1}}$$ which is zero by inspection.
Hint: Suppose $0 \leq x \leq 1$, and consider a coin with bias $x$ being flipped until one of its sides comes up $n+1$ times. The left-hand side counts the probability that this happens, which is plainly $1$. Since both sides are polynomials in $x$ and the identity is true for infinitely many values of $x$, it must be true for all $x$.