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$${n}\choose{r}$ = 0,
= 2020(0) + 1011${2020}\choose{1010}$
i.e. from the last term to the term in the middle I flipped the equation to equate the binomial coefficients.
However the number I got is so large my calculator can't process it so I am not sure how to proceed and if this method is correct.
Best Answer
This sum can be written as $$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 $S=0$.