# Counting summation of binomial coefficients

I need to count the summation of binomial coefficients in the sequence

$${2020}\choose{0}$$ - 2 $${2020}\choose{1}$$ + 3$${2020}\choose{2}$$ + ... - 2020$${2020}\choose{2019}$$ + 2021$${2020}\choose{2020}$$

i.e. all even numbered multipliers are negative, and sum is $$\sum^{2020}_{r=0} (-1)^{r} (r + 1) {2020}\choose{r}$$

I simplified this sequence using the symmetry rule to

$${2020}\choose{0}$$ - 2$${2020}\choose{1}$$ + 3$${2020}\choose{2}$$ + ... - 1010$${2020}\choose{2009}$$ + 1011$${2020}\choose{2010}$$ +

2021$${2020}\choose{0}$$ - 2020 $${2020}\choose{1}$$ + 2019$${2020}\choose{2}$$ + ... - 1012$${2020}\choose{2009}$$

= 2022$$\sum^{1009}_{r = 0} (-1)^r{2020}\choose{r}$$ + 1011$${2020}\choose{1010}$$

Since $$\sum^{n}_{r = 0} (-1)^r{n}\choose{r}$$ = 0,

= 2020(0) + 1011$${2020}\choose{1010}$$

i.e. from the last term to the term in the middle I flipped the equation to equate the binomial coefficients.

However the number I got is so large my calculator can't process it so I am not sure how to proceed and if this method is correct.

This sum can be written as $$S=\sum_{k=0}^{n} (-1)^k (k+1){n \choose k}~~~~(1)$$The binomial theorem : $$\sum_{k=0}^{n} {n \choose k} (-x)^k=(1-x)^n~~~(2)$$ It gives $$\sum_{k=0}^{n}(-1)^k {n \choose k}=0 ~~~(3)$$ D (1) w.r. t $$x$$ and put $$x=1$$, we again get $$\sum_{k=0}^{n} (-1)^k k {n \choose k}=0~~~~(4)$$ From (3,4) it follows that $$S=0$$.

$${(1+x)}^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2..\binom{n}{n}x^n$$

multiply x on both sides

$$x{(1+x)}^n=\binom{n}{0}x+\binom{n}{1}x^2+\binom{n}{2}x^3..\binom{n}{n}x^{n+1}$$

now you can replace x by -x

$$x{(1-x)}^n=\binom{n}{0}x-\binom{n}{1}x^2+\binom{n}{2}x^3..{(-1)}^n\binom{n}{n}x^{n+1}$$

now you can differentiate wrt x on both sides and put the value $$x=1$$ to get your sum (here n=2020)

I think you're on the right track. Note that you don't have $$\sum_{r=0}^{2020}(-1)^r\binom {2020}r$$, so you can't make that simplification. However, instead of replacing $$\binom nr$$ with $$\binom n{n-r}$$ for only half the list, what would happen if you tried the replacement for the entire sum, then add the original sum to the new sum?

Since this answer has more or less been ignored, let me make it clearer. You have already determined your sum

$$k=\sum_{r=0}^{2020}(-1)^r(r+1)\binom{2020}r$$

Now, use $$\binom{2020}r=\binom{2020}{2020-r}$$ and substitute $$u=2020-r$$ to get

$$k=\sum_{u=0}^{2020}(-1)^{2020-u}(2020-u+1)\binom{2020}u=\sum_{u=0}^{2020}(-1)^u(2021-u)\binom{2020}u$$

Of course, the index of the summation is just a placeholder. We can replace the $$u$$ with $$r$$ again and add term by term to get

$$2k=\sum_{r=0}^{2020}(-1)^r(2022)\binom{2020}r=2022(1-1)^{2020}=0$$

Define $$f(x,y) = x(x+y)^{2020}$$. Then the sum your looking to calculate is equal to $$\frac{\partial f}{\partial x}(1,-1)$$.