First slide

Binomial theorem for positive integral Index

Question

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 all $k \geq n$ , then b_n is equal to

Difficult

A

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

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

$^{2 n} C_{n}$
D

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

Solution

In the given equation, put x - 3 = y.
$∴ \sum_{r = 0}^{2 n} a_{r} (1 + y)^{r} = \sum_{r = 0}^{2 n} b_{r} (y)^{r}$
$\Rightarrow a_{0} + a_{1} (1 + y) + \dots + a_{n - 1} (1 + y)^{n - 1} + (1 + y)^{n} + (1 + y)^{n + 1} + \dots + (1 + y)^{2 n}$
$= \sum_{r = 0}^{2 n} b_{r} y^{r}$ $[Using a_{k} = 1, \forall k \geq n]$
Equating the coefficients of yⁿ on both sides, we get
$^{n} C_{n} +^{n + 1} C_{n} +^{n + 2} C_{n} + \dots +^{2 n} C_{n} = b_{n}$
$\Rightarrow (^{n + 1} C_{n + 1} +^{n + 1} C_{n}) +^{n + 2} C_{n} + \dots +^{2 n} C_{n} = b_{n}$
$[{Using}^{n} C_{n} =^{n + 1} C_{n + 1} = 1], [n_{c_{r}} + n_{c_{r - 1}} = {(n + 1)}_{c_{r}}]$
$\Rightarrow b_{n} =^{n + 2} C_{n + 1} +^{n + 2} C_{n} + \dots +^{2 n} C_{n}$
Combing the terms in similar way, we get
$\Rightarrow b_{n} =^{2 n} C_{n + 1} +^{2 n} C_{n} =^{2 n + 1} C_{n + 1}$

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