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#### Pranav

##### Well-known member

- Nov 4, 2013

- 428

**Question:**

If $\displaystyle \sum_{r=0}^{2n} a_r(x-2)^r=\sum_{r=0}^{2n} b_r(x-3)^r$ and $a_k=1$ for all $k \geq n$, then show that $b_n={}^{2n+1}C_{n+1}$.

**Attempt:**

I haven't been able to make any useful attempt on this one. I could rewrite it to:

$$\sum_{r=0}^{n-1} a_r(x-2)^r + (x-2)^n\left(\frac{(x-2)^{n+1}-1}{(x-2)-1}\right)=\sum_{r=0}^{2n} b_r(x-3)^r$$

I am clueless about the next step.

Any help is appreciated. Thanks!