Binomial Theorem 1 Question 31
33.
If $\sum_{r=0}^{2 n} a_{r}(x-2)^{r}=\sum_{r=0}^{2 n} b_{r}(x-3)^{r}$ and $a_{k}=1, \forall k \geq n$, then show that $b_{n}={ }^{2 n+1} C_{n+1}$
$(1992,6 \mathrm{M})$
Show Answer
Solution:
- Let $y=(x-a)^{m}$, where $m$ is a positive integer, $r \leq m$
Now, $\frac{d y}{d x}=m(x-a)^{m-1} \Rightarrow \frac{d^{2} y}{d x^{2}}=m(m-1)(x-a)^{m-2}$
$ \Rightarrow \quad \frac{d^{3} y}{d x^{3}}=m(m-1)(m-2)(m-3)(x-a)^{m-4} $
On differentiating $r$ times, we get
$ \begin{aligned} \frac{d^{r} y}{d x^{r}} & =m(m-1) \ldots(m-r+1)(x-a)^{m-r} \\ & \left.=\frac{m !}{(m-r) !}(x-a)^{m-r}=r !{ }^{m} C_{r}\right)(x-a)^{m-r} \end{aligned} $
and for $r>m, \frac{d^{r} y}{d x^{r}}=0$
Now, $\sum_{r=0}^{2 n} a_{r}(x-2)^{r}=\sum_{r=0}^{2 n} b_{r}(x-3)^{r}$
[given]
On differentiating both sides $n$ times w.r.t. $x$, we get
$ \sum_{r=n}^{2 n} a_{r}(n !)^{r} C_{n}(x-2)^{r-n}=\sum_{r=n}^{2 n} b_{r}(n !)^{r} C_{n}(x-3)^{r-n} $
On putting $x=3$, we get $\sum_{r=n}^{2 n} a_{r}(n !)^{r} C_{n}=\left(b_{n}\right) n$ !
[since, all the terms except first on RHS become zero]
$ \begin{aligned} & \Rightarrow \quad b_{n}={ }^{n} C_{n}+{ }^{n+1} C_{n}+{ }^{n+2} C_{n}+\ldots+{ }^{2 n} C_{n} \\ & \quad\left[\because a_{r}=1, \forall r \geq n\right] \\ &=\left({ }^{n+2} C_{n+1}+{ }^{n+2} C_{n}\right)+\ldots+{ }^{2 n} C_{n} \\ &={ }^{n+3} C_{n+1}+\ldots+{ }^{2 n} C_{n}=\ldots \\ &={ }^{2 n} C_{n+1}+{ }^{2 n} C_{n}={ }^{2 n+1} C_{n+1} \end{aligned} $
