If $\sum_{r = 0}^{2 n} a_{r} {(x - 2)}^{r} = \sum_{r = 0}^{2 n} b_{r} {(x - 3)}^{r}$ and $a_{k} = 1,$ for $k \geq n$ then $b_{n} =$

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$^{2 n} C_{n + 1}$

$^{2 n + 1} C_{n + 1}$

$^{2 n} C_{n - 1}$

$^{2 n + 2} C_{n}$

answer is B.

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Detailed Solution

   $∵ \sum_{r = 0}^{2 n} a_{r} {(x - 2)}^{r} = \sum_{r = 0}^{2 n} b_{r} {(x - 3)}^{r}$
Let    $y = x - 3 \Rightarrow y + 1 = x - 2$
So, the given expression reduces to
$\sum_{r = 0}^{2 n} a_{r} {(1 + y)}^{r} = \sum_{r = 0}^{2 n} b_{r} y^{r}$
$\Rightarrow a_{0} + a_{1} (1 + y) + a_{2} {(1 + y)}^{2} + .... + a_{2 n} {(1 + y)}^{2 n} = b_{0} + b_{1} y + ... + b_{2 n} y^{2 n}$
Using $a_{k} = 1, \forall k \geq n,$ we get
$a_{0} + a_{1} (1 + y) + a_{2} {(1 + y)}^{2} + .... + a_{n - 1} {(1 + y)}^{n - 1} + {(1 + y)}^{n} + {(1 + y)}^{n + 1} + .... + {(1 + y)}^{2 n}$
$= b_{0} + b_{1} y + .... + b_{n} y^{n} + ....... + b_{2 n} y^{2 n}$
On comparing the coefficient of $y^{n}$ on both sides, we get
$^{n} C_{n} +^{n + 1} C_{n} +^{n + 2} C_{n} + ..... +^{2 n} C_{n} = b_{n}$
$\Rightarrow^{n + 1} C_{n + 1} +^{n + 1} C_{n} +^{n + 2} C_{n} + ..... +^{2 n} C_{n} = b_{n}$
   $[∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}]$
$\Rightarrow^{n + 2} C_{n + 1} +^{n + 2} C_{n} + ..... +^{2 n} C_{n} = b_{n}$
[adding first two terms]
If we combine terms on LHS finally, we get
$^{2 n + 1} C_{n + 1} = b_{n}$