Find sum $S = {n \choose 1} – 2 {n \choose 2} + 3{n \choose 3} – 4{n \choose 4} + \cdots + (-1)^{n} n {n \choose n}$
I need verification of my results.
Assume that $n > 0$
So we see that $k {n \choose k} > 0$ for every odd $k$ , and $k {n \choose k} < 0$ for every even $k$.
However last term is defined as $(-1)^n n {n \choose n}$, which breaks the pattern, since it's $>0$ for even $n$, and $<0$ for odd $n$.
So I decided to define $S$ as follows:
$S = S + (-1)^{n-1} n {n \choose n} – (-1)^{n-1} n {n \choose n}$.
Obviously, $-(-1)^{n-1} n {n \choose n} = (-1)^n n {n \choose n}$.
So now $S$ looks like this:
$S = {n \choose 1} – 2 {n \choose 2} + 3{n \choose 3} – 4{n \choose 4} + \cdots + (-1)^{n-1} n {n \choose n} + (-1)^{n} 2n {n \choose n} = (-1)^{n} 2n {n \choose n} + \sum_{k = 1}^{n}(-1)^{k-1} k{n \choose k}$
By this results, we have:
$\sum_{k = 1}^{n}(-1)^{k-1} k{n \choose k} =\sum_{k = 1}^{n} (-1)^{k-1}n {n-1 \choose k-1} = n\sum_{k = 1}^{n} (-1)^{k-1} {n-1 \choose k-1}= n\sum_{k = 0}^{n-1} (-1)^{k}{n-1 \choose k}=$
$n\sum_{k = 0}^{n-1} (-1)^{k}1^{n-1-k
}{n-1 \choose k} = n(1-1)^{n-1} = 0$
Hence, $S = (-1)^{n} 2n {n \choose n}$.
Am I right? Did I overlook something? Posting this question since my mate somehow found this sum equal to $0$ for every $n$.
UPD.
Indeed, there was a typo. Not mine, but by problem setter. Thank you for your time!
Best Answer
If $n>1$ we have $$\sum_{k=0}^nC^k_ns^k=(1+s)^n\Rightarrow \sum_{k=1}^nkC^k_ns^{k-1}=n(1+s)^{n-1}\Rightarrow_{s={-1}} S=0$$