Power Series with $a_{n}$ and $a_{k}$ 
If $P(x)= \sum_{n=0}^{\infty} a_{n} x^{n}$. It is known that $P$ satisfies: $$P^{\prime}(x)=2xP(x)$$ $\\$ for all $ x\in \mathbb{R}$ and $$P(0)= 1$$
\begin{split}\end{split} (1) Prove that $a_{2k+1}=0$ for every $ k\in \mathbb{\left \{1,2,3,4,...  \right \}}$

Its true that if two power series have the same coefficients then they are equal?
I can demonstrate the first equation transforming the series to an integral, but i cant prove (1) and (2) because $k$ confuses me, I was thinking that $k$ is for $P'(x)$, just as $n$ is for $P(x)$. Also i need orientation with the proof requested.
Thanks.
 A: It is true that if two power series have the same coefficients then they are equal. More is true actually. If two power series are equal on some interval then their coefficients must be equal.
Assuming that the series has a positive radius of convergence $R$, we can differentiate term by term to obtain a series for $P'(x)$.
$$P'(x)=\sum_{k=1}^{\infty}{ka_kx^{k-1}}$$
This series also has radius of convergence $R$, and since $P'(x)=2xP(x)\implies P'(0)=a_1=0$
Also, since $P'(x)=2xP(x)$
$$\sum_{k=1}^{\infty}{ka_kx^{k-1}}=\sum_{k=0}^{\infty}{2a_kx^{k+1}}$$
making a shift of index we see that,
$$\sum_{k=1}^{\infty}{(k+1)a_{k+1}x^k}=\sum_{k=1}^{\infty}{2a_{k-1}x^k}$$
Since these power series are equal on some interval with radius $R>0$, their coefficients must be equal. Thus we have,
$$(k+1)a_{k+1}=2a_{k-1}$$
Now you can probably finish the proof on your own using induction.
A: If $P(x)$ is of degree $n$, then the lhs is of degree $(n-1)$ and the rhs is of degree $(n+1)$.
Since we compare the coefficients of same power, we  have only odd or only even powers.
Now, use the fact that for power $0$, $P(0)\neq 0$ and you have the answer.
