Alternating sum of binomial of coefficients up to $\binom{n}{r}$

Question

Question : Evaluate - $$ \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r}$$

I tried to solve it but I am able to solve only when $r=n$.In that case I can put $x=-1$ in the expansion of $(1+x)^n$ and the result is $0$.

I know that for asking any question on this forum, I need to show my approach and thoughts but literally I have no idea how to solve up to $r+1$ terms.

Although I tried to add and subtract the terms to reach upto $\binom{n}{n}$, and evaluate the remaining sum, but nothing helpful.

Any help will be appreciable!

Answer 1

You have

$${n \choose k}={n-1 \choose k-1}+{n-1\choose k}$$ meaning that to get the next line in the Pascal's triangle, you sum two numbers above. Now you're computing an alternating sum, so you get a telescoping effect:

$${n\choose 0}-{n\choose 1}+\cdots+(-1)^r{n\choose r}=$$ $$=\bigg(0+{n-1\choose 0}\bigg)-\bigg({n-1\choose 0}+{n-1\choose 1}\bigg)+\cdots + (-1)^r\bigg({n-1 \choose r-1}+{n-1\choose r}\bigg)=$$ $$\textstyle=\big[{n-1\choose 0}-{n-1\choose 0}\big]+\big[{n-1\choose 1}-{n-1\choose 1}\big]+\cdots + \big[{n-1 \choose r-1}-{n-1\choose r-1}\big]+(-1)^r{n-1\choose r}=$$ $$=(-1)^r {n-1\choose r}$$

Short answer: you're counting each number from the upper row of the Pascal's triangle twice with opposite signs, except from the $r$-th one (including the sign).

Answer 2

$$\sum_{k=0}^r (-1)^k {n \choose k} = (-1)^r {n-1 \choose r} $$

Answer 3

I also apply: $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}$$ But in another way, call: $$f(n) = \binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\binom{n}{3} \dots + (-1)^r\binom{n}{r} = \sum\limits_{k=0}^{r} \binom{n}{k}-1^k$$

$$f(n+1)-f(n)=\sum\limits_{k=0}^{r} -1^k (\binom{n+1}{k}-\binom{n}{k})=\sum\limits_{k=0}^{r} -1^k \binom{n}{k-1}=\sum\limits_{k=0}^{r-1} -1^{k+1} \binom{n}{k}$$

$$f(n+1)=f(n+1)-f(n)+f(n)=\sum\limits_{k=0}^{r-1} -1^{k+1} \binom{n}{k} + \sum\limits_{k=0}^{r} \binom{n}{k}-1^k = -1^r\binom{n}{r} + \sum\limits_{k=0}^{r-1} (-1^k+(-1^{k+1})) \binom{n}{k} = -1^r\binom{n}{r}$$

Because $$f(n+1) = -1^r\binom{n}{r}$$ So $$f(n) = -1^r\binom{n-1}{r}$$