SATHEE: Binomial Theorem 1 Question 31

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

$(1992, 6 M)$

Show Answer

Solution:

Let $y = (x - a)^{m}$ , where $m$ is a positive integer, $r \leq m$

Now, $\frac{d y}{d x} = m (x - a)^{m - 1} \Rightarrow \frac{d^{2} y}{d x^{2}} = m (m - 1) (x - a)^{m - 2}$

$\Rightarrow \frac{d^{3} y}{d x^{3}} = m (m - 1) (m - 2) (m - 3) (x - a)^{m - 4}$

On differentiating $r$ times, we get

$\begin{aligned} \frac{d^{r} y}{d x^{r}} & = m (m - 1) \dots (m - r + 1) (x - a)^{m - r} \\ = \frac{m!}{(m - r)!} (x - a)^{m - r} = r!^{m} C_{r}) (x - a)^{m - r} \end{aligned}$

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 $\sum_{r = n}^{2 n} a_{r} (n!)^{r} C_{n} = (b_{n}) n$ !

[since, all the terms except first on RHS become zero]

$\begin{aligned} \Rightarrow b_{n} =^{n} C_{n} +^{n + 1} C_{n} +^{n + 2} C_{n} + \dots +^{2 n} C_{n} \\ [∵ a_{r} = 1, \forall r \geq n] \\ = (^{n + 2} C_{n + 1} +^{n + 2} C_{n}) + \dots +^{2 n} C_{n} \\ =^{n + 3} C_{n + 1} + \dots +^{2 n} C_{n} = \dots \\ =^{2 n} C_{n + 1} +^{2 n} C_{n} =^{2 n + 1} C_{n + 1} \end{aligned}$

Binomial Theorem 1 Question 31

33.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Binomial Theorem 1 Question 31

33.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu