Based on the binomial expansion of $(1+x)^n$, show that:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{n + k}{k} = (-1)^n.$$
This is a question from a very old high school exam paper I came across. It looks like it involes the product of two binomial expansions but I cannot figure this out. Any help much appreciated.
If you write the equation to prove as $\sum_k(-1)^{n-k}\binom nk\binom{n+k}n=1$, you can interpret the left hand side as the result of applying the $n$-th finite difference operation $\Delta^n$ to the sequence $k\mapsto\binom kn$, and taking the term of the resulting sequence at $k=n$. Then $\Delta(k\mapsto\binom km)=k\mapsto\binom k{m-1}$ implies that the left hand side becomes $\Delta^n\left(k\mapsto\binom kn\right)(n)=\binom n{n-n}=\binom n0=1$ as desired.
Another approach is to recognise negation of the upper index in the second binomial coefficient: $$ (-1)^k\binom{n+k}k=\binom{k-(n+k)-1}k=\binom{-n-1}k, $$ after which the identity becomes in instance of the Vandermonde identity $$ \sum_{k=0}^n(-1)^k\binom nk\binom{n + k}k= \sum_k\binom n{n-k}\binom{-n-1}k=\binom{-1}n=(-1)^n. $$
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Added after request in comment. Neither of these arguments requires much beyond basic knowledge about binomial coefficients. The difference operator $\Delta$ is defined by $\Delta(f)(k)=f(k+1)-f(k)$, so with $f:k\mapsto\binom km$ one gets $$ \Delta(f)(k)=\binom{k+1}m-\binom km=\binom k{m-1} $$
just from Pascal's recurrence. And the formula $$ \Delta^n(f)(n)=\sum_i(-1)^{n-i}\binom nif(n+i) $$ either follows easily by induction from the definition of $\Delta$, or if you prefer by using the operators identity $I$ and shift $S$ on functions, defined by $I(f)=f$ and $S(f)(i)=f(i+1)$, writing $\Delta=S-I$, and using the binomial formula for $(S-I)^n$ (possible since $S$ and $I$ commute).
In the second variant, you probably already know the formula $\binom{-n}k= (-1)^k\binom{n+k-1}k$ for binomial coefficients with negative upper index, or else the equality between the outer expressions in $$ (1+X)^{-n} = \sum_{k\in\mathbf N} \binom{-n}kX^k = \sum_{k\in\mathbf N}(-1)^k\binom{n+k-1}kX^k $$ (if not, see this answer deriving the first formula). Then $\sum_k(-1)^k\binom{n+k}kX^k=(1+X)^{-n-1}$, and the instance of the Vandermonde identity is proved by comparing coefficients in $$ \sum_n\left(\sum_{k=0}^n(-1)^k\tbinom{n + k}k\tbinom nk\right)X^n =(1+X)^{-n-1}(1+X)^n=(1+X)^{-1}=\sum_n(-1)^nX^n $$ which might be the proof using only manipulation of $(1+X)^{n-1}$ you asked for.
Suppose we seek to evaluate $$\sum_{k=0}^n (-1)^k {n\choose k} {n+k\choose k}.$$
Start from $${n+k\choose k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{k+1}} (1+z)^{n+k} \; dz.$$
This yields the following expression for the sum $$\frac{1}{2\pi i} \int_{|z|=\epsilon} \sum_{k=0}^n {n\choose k} \frac{(-1)^k}{z^{k+1}} (1+z)^{n+k} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \sum_{k=0}^n {n\choose k} (-1)^k \frac{(1+z)^k}{z^k}\; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \left(1-\frac{z+1}{z}\right)^n \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z} \frac{(-1)^n}{z^n} \; dz \\ = \frac{(-1)^n}{2\pi i} \int_{|z|=\epsilon} \frac{(1+z)^n}{z^{n+1}}\; dz.$$
It follows that the closed form of the sum is given by $$(-1)^n [z^n] (1+z)^n = (-1)^n.$$
A trace as to when this method appeared on MSE and by whom starts at this MSE link.
The following combinatorial proof may be of interest as well. After multiplying both sides by $(-1)^n$, and reindexing the summation by replacing $k$ with $n-k$, we can rewrite the equation to be proved as $$ \sum_{k=0}^n(-1)^k\binom{n}k\binom{2n-k}n=1 $$ Both sides are answer to the following question:
How many strings of $n$ ones and $n$ zeroes are there were no zero is immediately followed by a one?
Obviously, there is only one such string, $11\cdots100\cdots0$.
On the other hand, we can count this using the principle of inclusion exclusion. Take all of the $\binom{2n}n$ strings of $n$ zeroes and $n$ ones, and for each $i=1,2,\dots,n$, subtract the strings where the $i^{th}$ zero is followed by a ones. Every such string can be generated by taking an arbitrary string of $n$ zeroes and only $n-1$ ones, then adding a one after the $i^{th}$ zero. Therefore, for each $i=1,\dots,n$, we must subtract $\binom{2n-1}n$, so subtract $n\binom{2n-1}n$.
However, strings with two instances of a zero followed by a one have been double subtracted, so these must be added back in. Using a similar method, the number strings where both the $i^{th}$ zero and the $j^{th}$ zero are followed by ones is $\binom{2n-2}n$. We must add this for each of the $\binom{n}2$ pairs of zeroes, so add in $\binom{n}2\binom{2n-2}n$.
Continuing in this fashion (subtracting in the triply subtracted strings, adding in the quadruply subtracted strings, etc), we recover the alternating sum on the left hand side.
Just for seeing a direct method:
$$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\binom{n + k}{k}= \sum_{k=0}^{n}\binom{n}{n-k}\binom{-n-1 }{k} $$$$=\binom{-1 }{n}=\left(-1\right)^{n}\binom{n}{n}=\left(-1\right)^{n}$$
