<s>[INST] 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?
 [/INST] 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 $${2n+1 \choose n+1}=B_n$$. 
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