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 $b_n=\displaystyle{2n+1\choose n+1}$

Question

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?

Answer 1

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 $a_k=1$.

Answer 2

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*}

Answer 3

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 $${2n+1 \choose n+1}=B_n$$.

Answer 4

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