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 $b _n={ }^{2 n+1} C _{n+1}$
$(1992,6 M)$
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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]
$$ \begin{aligned} & \Rightarrow \quad b _n={ }^{n} C _n+{ }^{n+1} C _n+{ }^{n+2} C _n+\ldots+{ }^{2 n} C _n \\ & \quad\left[\because a _r=1, \forall r \geq n\right] \\ &=\left({ }^{n+2} C _{n+1}+{ }^{n+2} C _n\right)+\ldots+{ }^{2 n} C _n \\ &={ }^{n+3} C _{n+1}+\ldots+{ }^{2 n} C _n=\ldots \\ &={ }^{2 n} C _{n+1}+{ }^{2 n} C _n={ }^{2 n+1} C _{n+1} \end{aligned} $$
