A proof in combinatorics (prove by any method) I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following by using any method?
$$\left( \begin{array}{c} n \\ 1\ \end{array} \right) + 3 \left( \begin{array}{c} n \\ 3\ \end{array} \right) +5 \left( \begin{array}{c} n \\ 5\ \end{array} \right) +... = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + 4 \left( \begin{array}{c} n \\ 4\ \end{array} \right)+... $$
Could use some help as I work through possible exam questions. 
 A: Putting the right-hand side to the left we have
\begin{align*}
  \binom{n}{1}-2\binom{n}{2}+3\binom{n}{3}-\cdots=
  \begin{cases}
    1\qquad n=1\\
    0\qquad n>1
    \end{cases}
  \end{align*}
or in more compact notation using Iverson brackets:
\begin{align*}
  \sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}=[[n=1]]
  \end{align*}

We obtain for integer $n>0$
  \begin{align*}
\color{blue}{\sum_{k=1}^n(-1)^{k+1}k\binom{n}{k}}
&=n\sum_{k=1}^n(-1)^{k+1}\binom{n-1}{k-1}\tag{1}\\
&=n\sum_{k=0}^{n-1}(-1)^k\binom{n-1}{k}\tag{2}\\
&=n(1-1)^{n-1}\tag{3}\\
&\color{blue}{=[[n=1]]}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we apply the binomial identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$.

*In (2) we shift the index $k$ to start with $k=0$.

*In (3) we apply the binomial theorem.
A: The way @EwanDelanoy has told in the comments is absolutely correct.
I shall be using $C_0$, $C_1$ etc. for $\binom{n}{0}$, $\binom{n}{1}$ and so on.
Here's how the proof goes : 
$$(1 - x)^n = C_0 - C_1x + C_2x^2 - C_3x^3 + ..... + (-1)^n.C_nx^n$$

Differentiating both sides with respect to $x$,
$$-n(1 - x)^{n-1} = -C_1 + 2.C_2x - 3.C_3x^2 + 4.C_4x^3 - ....$$ (You can verify this by using the Chain Rule) 

On putting $x = 1$,
$$0 = -C_1 + 2.C_2 - 3.C_3 + 4.C_4 - ....$$
$$\implies C_1 + 3.C_3 + 5.C_5 + .... = 2.C_2 + 4.C_4 + 6.C_6 + ....$$ (Taking the negative terms to the L.H.S.)
