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Chapter 7: Problem 583

If co-efficient of \((\mathrm{r}+2)\) th term and \(3 \mathrm{r}\) th term are equal in expansion of \((1+\mathrm{x})^{2 \mathrm{n}}, \mathrm{n}, \mathrm{r} \in \mathrm{N}, \mathrm{r}>1, \mathrm{n}>2\) then \(\mathrm{n}=\) (a) \(3 \mathrm{r}\) (b) \(3 \mathrm{r}+1\) (c) \(2 \mathrm{r}\) (d) \(2 r+1\)

Short Answer

Expert verified
The value of n is \(n = 2r+1\) (Option d).

Step by step solution

01

Binomial Coefficient Formula

The general formula for the nth term of a binomial expansion is: \[\binom{n}{r} = \frac{n!}{r!(n - r)!}\] In this problem, we are using the binomial expansion of (1+x)^{2n}, so the coefficient of the rth term can be written as: \[\binom{2n}{r}\]
02

Compare the coefficients

Given that the coefficients of (r+2)th term and 3rth term are equal. We can write this as: \[\binom{2n}{r+2} = \binom{2n}{3r}\]
03

Apply Binomial Coefficient Formula

Apply the binomial coefficient formula to both sides of the equation and simplify: \[\frac{(2n)!}{(r+2)!(2n-(r+2))!} = \frac{(2n)!}{(3r)!(2n-3r)!}\]
04

Cancel out common factors

As both sides of the equation share a common factor of (2n)!, we can cancel it out: \[\frac{1}{(r+2)!(2n-(r+2))!} = \frac{1}{(3r)!(2n-3r)!}\] Now, we can write the above equation as: \[(r+2)!(2n-(r+2))! = (3r)!(2n-3r)!\]
05

Solve for n

To find the value of n, we use the given constraints: n, r ∈ N, r > 1 and n > 2. For r > 1, let's first try r = 2: \[(2+2)!(2n-(2+2))! = (3\cdot2)!(2n-3\cdot2)!\] \[(4!)(2n-4)! = (6!)(2n-6)!\] Now let's consider the smallest possible value of n > 2 i.e., n = 3: \[(4!)(2\cdot3-4)! = (6!)(2\cdot3-6)!\] This simplifies to: \[4! = 6!\] Which is clearly not true. Thus, in order to satisfy the given constraints, the value of n must be:
06

Answer:

\[\boxed{n = 2r+1}\] (Option d)

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