How is ${{-2}\choose {n}}=(-1)^n(n+1)$ true?

Question

Show that $$\frac{1}{(1-z)^2}=\sum_\limits{n=0}^{\infty}(n+1)z^n1\:,\:|z|<1$$

It was suggested in the book that the reader should use the taylor expansion of the following function $(1+z)^{\alpha}=1+\sum_\limits{n=0}^{\infty}{{\alpha}\choose n}z^n$

However ${{-2}\choose {n}}=\frac{-2!}{(-2-n)!n!}$ but on the book it is stated instead ${{-2}\choose {n}}=(-1)^n(n+1)$.

Questions:

How can ${{-2}\choose {n}}=(-1)^n(n+1)$? Is ${{-2}\choose {n}}$ the combinations notation? Or does it stand for another concept?

Thanks in advance!

Answer 1

$$\binom{-2}{n}=\frac{(-2)(-3)...(-1-n)}{n!} = (-1)^n \frac{(2)(3)...(n+1)}{n!}=(-1)^n\frac{(n+1)!}{n!}=(-1)^n(n+1)$$

Answer 2

Hint: It is based on the general definition $$\binom{\alpha}{n} = \frac{\prod\limits _{j=0}^{n-1}(\alpha -j)}{n!},$$ where $n$ is a non-negative integer and $\alpha$ is any real or complex number. So

$$\begin{align} \binom{-2}{n} &= \frac{\prod\limits _{j=0}^{n-1}(-2 -j)}{n!}\\ &= \frac{(-1)^n\prod\limits _{j=0}^{n-1}(2 +j)}{n!}\quad (\text{remember in general }\prod\limits_{j=0}^{n-1}\color{blue}{c}a_j = \color{blue}{c^n}\prod\limits_{j=0}^{n-1}a_j ).\end{align}$$

Can you finish from here?

Answer 3

It is similar to the combination notation that you would have first learned in an elementary combinatorics class which is used for counting problems but is subtly different.

The first definition you would have likely learned for $\binom{n}{k}$ is as it being the answer to the question "In how many ways may we select $k$ objects out of $n$ where order of selection doesn't matter?" We learn that $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$. This definition of course makes little to no sense in the case where $n$ and $k$ are not non-negative integers. How can you "select half of an object" or "select two objects out of $-\pi$" etc...

Despite the limitation of the definition first learned in an elementary counting setting, the binomial coefficient proved to be terribly useful in establishing algebraic identities such as the well known $(a+b)^n = \sum\limits_{i=0}^n \binom{n}{i}a^ib^{n-i}$

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In an attempt to make it even more convenient for algebra, the definition can be slightly altered to instead be $\binom{n}{k} = \dfrac{n\frac{k}{~}}{k!}$ where here $n\frac{k}{~}$ represents the falling factorial $\underbrace{n(n-1)(n-2)\cdots (n-k+1)}_{k~\text{terms}}$. In doing so, the new extended definition gives the same result for all cases for $n$ being a non-negative integer, but we can now assign values to things where $n$ is no longer non-negative or even when $n$ is not an integer at all. ($k$ must still be a non-negative integer for the formula to make sense).

In doing so, we can now have a more expansive binomial theorem: $(1+x)^\alpha = \sum\limits_{i=0}^\infty \binom{\alpha}{i}x^i$ where this now works even for non-positive or even non-integer exponents. Note further that this actually agrees with the earlier version of the binomial theorem in the case that $\alpha$ is a positive integer (as eventually the falling factorials in the numerator will become zero).

All of the identities learned for binomial coefficients previously still apply for these newer extended binomial coefficients (apart from being able to automatically say that the result is zero for some situations) and there are additional identities that you can discover with them. In particular $\binom{n}{k}=(-1)^k\binom{-n+k-1}{k}$ which is a more general result than the one you point to in your question.