Expand the series and find the summation. 
$f(x)=\sum_{k=0}^{n}a_{k}x^{k}$

find $f(2x+1)$
Is this the correct expansion

$f(2x+1)=\sum_{k=1}^na_k\sum_{i=1}^k{k\choose i}(2x)^i$
Is it possible to do the following step:
  

 A: The question is not clear to me, but I thought you might want to have a polynomial in $x$ for example of the form $$f(2x+1)=\sum_{i=0}^nb_ix^i.$$ If this is the case then proceed as follows:
\begin{align}
f(2x+1)&=\sum_{k=0}^na_k\sum_{i=0}^k{k\choose i}(2x)^i\\
&=\sum_{i=0}^n\sum_{k=i}^na_k{k\choose i}(2x)^i\\
&=\sum_{i=0}^n\Big[2^i\sum_{k=i}^na_k{k\choose i}\Big]x^i\\
&=\sum_{i=0}^nb_ix^i\\
\end{align}
where $b_i=2^i\sum_{k=i}^na_k{k\choose i}$.
A: The Binomial Theorem states: $(a+b)^n = \sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$
Applying this to your problem gives
$f(2x+1) = \sum_{k=0}^na_k(2x+1)^k = \sum_{k=0}^na_k\sum_{i=0}^k\binom{k}{i}(2x)^{k-i}1^i = \sum_{k=0}^na_k\sum_{i=0}^k\binom{k}{i}(2x)^i1^{k-i}$ (the last expression follows from the previous one by symmetry of the summands in the argument, as addition is commutative).
Of course, you needn't bother writing the "$1^i$" or "$1^{k-i}$" bits; those just there for demonstration, to show clearly how the Binomial Theorem is applied.
A: On base of $\left(a+b\right)^{k}=\sum_{i=0}^{k}\binom{k}{i}a^{i}b^{k-i}$
we find:
$f\left(2x+1\right)=\sum_{k=0}^{n}a_{k}\left(2x+1\right)^{k}=\sum_{k=0}^{n}a_{k}\sum_{i=0}^{k}\binom{k}{i}\left(2x\right)^{i}1^{k-i}=\sum_{k=0}^{n}a_{k}\sum_{i=0}^{k}\binom{k}{i}\left(2x\right)^{i}$
You could go further to achieve:
$=\sum_{k=0}^{n}\sum_{i=0}^{k}2^{i}a_{k}\binom{k}{i}x^{i}=\sum_{i=0}^{n}\sum_{k=i}^{n}2^{i}a_{k}\binom{k}{i}x^{i}=\sum_{i=0}^{n}b_{i}x^{i}$
where $b_{i}=2^{i}\sum_{k=i}^{n}a_{k}\binom{k}{i}$
