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?
 A: 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*}
A: Let $(x-3)=y$ Then we get
$$\sum_{r=0}^{2m} A_r(1+y)^r=\sum_{r-0}^{2n} B_r~ y^r$$
Compare the coefficient of $y^n$ both sides, then
$${n \choose n}+{n+1 \choose n}+{n+2 \choose n}+....+{2n \choose n}=B_n$$ or
$${n+1 \choose n+1}+{n+1 \choose n}+{n+2 \choose n}+....+{2n \choose n}$$
Next use $${n \choose r}+{n \choose r-1}={n+1 \choose r}$$ successively to get
The LHS in above as $${2n+1 \choose n+1}=B_n$$. 
A: Hint Derivate $n$ times and plug in $r=3$.
Let $f(x)=b_0+b_1(x-3)+b_2(x-3)^2+\cdots b_{2n}(x-3)^{2n}$. Then 
$$f^{(n)}(3)=n! b_n$$
Since 
$$f(x)=a_0+a_1(x-2)+a_2(x-2)^2+\cdots a_{2n}(x-2)^{2n}$$
you also have 
$$f^{(n)}{3}=n!a_n+\frac{(n+1)!}{1!}a_{n+1}+\frac{(n+2)!}{2!}a_{n+2}+...+\frac{(2n)!}{n!}a_{n+1}$$
Now make those equal and plug in $a_k=1$.
A: Write $x-3=y$
$$\sum_{r=0}^{2n}a_r(1+y)^r$$
$$=\sum_{r=0}^{2n}a_r(\sum_{k=0}^r\binom rky^k)$$
$$=\sum_{r=0}^{2n} \left(a_0+a_1(1+y)+a_2(1+\binom21y+y^2)+a_3(1+\binom31y+\binom32+y^3)+\cdots+a_{2n}(\cdots)\right)$$
So, the coefficient$(c_k)$ of $y^k$ will be $$\sum_{r=k}^{2n}a_r\binom rk$$
$$c_n=1+\binom{n+1}1+\binom{n+2}2+\cdots+\binom{2n}n$$
See https://en.m.wikipedia.org/wiki/Hockey-stick_identity
