If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$ If $\sum_{0}^{2n}a_r(x-2)^r=\sum_{0}^{2n}b_r(x-3)^r$ and $a_k=1$ $\forall$ $k\ge n$, then show that $b_n=\displaystyle{2n+1\choose n+1}$
My attempt is as follows:-
$$a_0+a_1(x-2)+a_2(x-2)^2+\cdots a_{2n}(x-2)^{2n}=b_0+b_1(x-3)+b_2(x-3)^2+\cdots b_{2n}(x-3)^{2n}$$
Let's see the term on R.H.S with coefficient as $b_n$
$$T_{n+1}=b_n(x-3)^n$$
Comparing the coefficients of $x^0$
$$a_0-2a_1+(-2)^2a_2+(-2)^3a_3+\cdots a_{2n}(-2)^{2n}=b_0-3b_1+(-3)^2b_2+(-3)^3b_3+\cdots+b_n(-3)^n+\cdots b_{2n}(-3)^{2n}$$
Comparing the coefficients of $x^1$
$$a_1+2a_2(-2)^1+3a_3(-2)^2\cdots 2n\cdot a_{2n}(-2)^{2n-1}=b_1+2a_2(-3)^1+3a_3(-3)^2\cdots nb_n(-3)^{n-1} \cdots (2n)a_{2n}(-3)^{2n-1}$$
Now I am not getting any way to find $b_n$ , its so mixed up in these expressions, not able to get any breakthrough. Any inputs?

We have
\begin{eqnarray*}
\sum_{r=0}^{2n} (x-2)^r = \sum_{r=0}^{2n} b_r (x-3)^r.
\end{eqnarray*}
Expand the $(x-2)$
\begin{eqnarray*}
(x-2)^r=(x-3+1)^r= \sum_{s=0}^{r} \binom{r}{s} (x-3)^s.
\end{eqnarray*}
Invert the order of the plums 
\begin{eqnarray*}
\sum_{r=0}^{2n} \sum_{s=0}^{r} \cdots = \sum_{s=0}^{2n} \sum_{r=s}^{2n} \cdots. 
\end{eqnarray*}
So we need to perform the following sum, which is the hockey stick identity 
\begin{eqnarray*}
b_s= \sum_{r=s}^{2n} \binom{r}{s} =\binom{2n+1}{s+1}. 
\end{eqnarray*}
In particular for $b_n$ we have
\begin{eqnarray*}
 b_n =\binom{2n+1}{n+1}. 
\end{eqnarray*}
