Simplifying $S=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...+(-1)^r\binom{n}{r}$

Question

Assume $n$ and $r$ are nonnegative integers where $$0\le r\le n$$How can one simplify $$S=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...+(-1)^r\binom{n}{r}$$ ?

I think that this combinatorial expression should be either solved using a combinatorial way but I don't know how. Probably It's also useful one tries to represent somehow using binomial representation of $(a+b)^n$. I appreciate any idea given!

Answer 1

Coefficient of $x^r$ in

$$(1-x)^n (1 + x + x^2 + ...) = (1-x)^{n-1}$$

So for $r \lt n$ it is $$(-1)^r\binom{n-1}{r}$$

Answer 2

Use the remainder term of a truncated Taylor series $$(1+x)^n=\sum_{k=0}^n{n \choose k}x^k=\sum_{k=0}^r{n \choose k}x^k + \int_0^x \frac{f^{(r+1)}(t)}{r!}(x-t)^r\,dt$$ Rearrange to put the truncated series on the LHS: $$\begin{align} \sum_{k=0}^r{n \choose k}x^k &= (1+x)^n - \int_0^x \frac{f^{(r+1)}(t)}{r!}(x-t)^r\,dt \\&= (1+x)^n - \frac{1}{r!}\frac{n!}{(n-r-1)!} \int_0^x (1+t)^{n-r-1}(x-t)^r\,dt \end{align}$$ Substitute $x=-1$: $$\begin{align} \sum_{k=0}^r{n \choose k}(-1)^k &= 0-\frac{1}{r!}\frac{n!}{(n-r-1)!} (-1)^{r}\int_0^{-1} (1+t)^{n-1}\,dt\\ &=\frac{(-1)^{r+1}}{r!}\frac{(n-1)!}{(n-r-1)!}[(1+t)^n]_0^{-1} \\&=\frac{(-1)^{r}}{r!}\frac{(n-1)!}{(n-r-1)!} \\&=(-1)^r{n-1\choose r} \end{align}$$

Answer 3

If you have no better idea, just work out the first terms and see what happens. So your first terms are:

$S_0=1$

$S_1=1-n$

$S_2=1-n+\cfrac {n(n-1)}2=\cfrac {n^2-3n+2}2=\cfrac {(n-1)(n-2)}2$

$S_3=\cfrac {(n-1)(n-2)}2-\cfrac{n(n-1)(n-2)}6=-\cfrac{(n-1)(n-2)(n-3)}6$

After which it is very clear what is happening and how this will go through easily by induction.

Answer 4

Just to present an alternative way to derive the result already provided in the answer by Aryabhata $$ \eqalign{ & S(r) = \sum\limits_{0\, \le \,k\, \le \,r} {\left( { - 1} \right)^{\,k} \left( \matrix{ n \cr k \cr} \right)} = \cr & = \sum\limits_{0\, \le \,k\, \le \,r} {\left( \matrix{ k - n - 1 \cr k \cr} \right)} = \quad \quad \quad (1) \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( \matrix{ r - k \cr r - k \cr} \right)\left( \matrix{ k - n - 1 \cr k \cr} \right)} = \quad \quad \quad (2) \cr & = \left( \matrix{ r - n \cr r \cr} \right) = \quad \quad \quad (3) \cr & = \left( { - 1} \right)^{\,r} \left( \matrix{ n - 1 \cr r \cr} \right)\quad \quad \quad (4) \cr} $$ where the following identities have been applied
(1) "Trinomial Revision";
(2) bounds on $k$ replaced by the 1st binomial;
(3) "Double convolution";
(4) "Trinomial Revision".