How to prove this result on binomial coefficients? Question: If $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\dotsm+\binom{n}{n}x^n$, prove that $\binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-\dotsm+(-1)^{n-1}n\binom{n}{n}=0$
My attempt:  I wrote the expression for $\binom{n}{k}$ for each term in the L.H.S and then tried to simplify but could not reach the R.H.S.
Please help. I have no idea on how to begin. I am not allowed to use calculus.
 A: $$\sum_{j \mathop = 1}^n \left({-1}\right)^{n + 1} j \binom n j=\sum_{j \mathop = 1}^n \left({-1}\right)^{n + 1} n \binom {n - 1} {j - 1}= n \sum_{j \mathop = 0}^{n - 1} \left({-1}\right)^{n - 1} \binom {n - 1} j=0$$
NOTE
we have used that
$$j \binom n j= n \binom {n - 1} {j - 1}$$
which has a simple explanation: select $j$ people out of n, then designate one as special. The LHS represents how many ways we can do this by first picking the $j$ people and then making designation. On the RHS, we have the number of ways to select the special one and then picking the remaining $j-1$ from the remaining $n-1$.
A: Hint: Replace ${ n\choose i}$ with $ { n-1 \choose i } + { n-1 \choose i+1 } $, and apply the condition that they provided.
$$ \begin{align} & \sum_{i=0}^n  (-1)^i i { n \choose i } \\
= & \sum_{i=0}^n ( -1)^i i \left[ { n-1 \choose i } + { n-1 \choose i+1 } \right]  \\
= & \sum_{i=0}^n {n-1 \choose i } \times \left[ (-1)^i i + (-1)^{i-1} (i-1) \right]  \\
= & \sum_{i=0}^n {n-1\choose i } (-1)^i \\
 =& ( 1 -1 ) ^{n-1} \\
= & 0 \end{align} $$

Notes:

*

*You'd have to check the indices, and ensure that they work out.

*The identity is true even if $ n-1 > i+1$, in which case $ { n-1 \choose i+1 } = 0 $.

