If `sum_(r=0)^(2n)a_r(x-2)^r=sum_(r=0)^(2n)b_r(x-3)^ra n da_k=1` for all `kgeqn ,` then show that `b_n=^(2n+1)C_(n+1)` .

Explanation

To solve the problem, we start with the given equation:

\[ \sum_{r=0}^{2n} a_r (x-2)^r = \sum_{r=0}^{n} b_r (x-3)^r \]

and the condition that \( a_k = 1 \) for all \( k \geq n \).

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