The coefficients of the $(3 r)^{t h}$ term and the $(r + 2)^{t h}$ term in the expansion $(1 + x)^{2 n}$ are equal, then :

n = 2 r

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n = 3 r

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n = 2 r + 1

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none of these

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Solution

The correct option is A $n = 2 r$
The coefficients of $(3 r)$ and $(r + 2)$ terms in the expansion $(1 + x)^{2 n}$ are same.
So, $^{2 n} C_{(3 r - 1)} =^{2 n} C_{(r + 1)}$
As we know that in general, $^{n} C_{r} =^{n} C_{n - r},$
$\Rightarrow 3 r - 1 + r + 1 = 2 n$
i.e. $4 r = 2 n$ or $n = 2 r$

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Given positive integers r > 1, n > 2 and the coefficients of $(3 r)^{t h}$ and $(r + 2)^{t h}$ terms in the Binomial expansion of $(1 + x)^{2 n}$ are equal, then:

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(r > 1, n > 2)

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and the coefficients of

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Q. Given positive integers

r > 1, n > 2

and the coefficient of

(3 r)^{t h}

and

(r + 2)^{t h}

terms in the binomial expansion of

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The coefficients of the (3r)th term and the (r+2)th term in the expansion (1+x)2n are equal, then :

The correct option is A n=2rThe coefficients of (3r) and (r+2) terms in the expansion (1+x)2n are same.So, 2nC(3r−1)=2nC(r+1)As we know that in general, nCr=nCn−r,⇒3r−1+r+1=2ni.e. 4r=2n or n=2r

The coefficients of the $(3 r)^{t h}$ term and the $(r + 2)^{t h}$ term in the expansion $(1 + x)^{2 n}$ are equal, then :

The correct option is A $n = 2 r$
The coefficients of $(3 r)$ and $(r + 2)$ terms in the expansion $(1 + x)^{2 n}$ are same.
So, $^{2 n} C_{(3 r - 1)} =^{2 n} C_{(r + 1)}$
As we know that in general, $^{n} C_{r} =^{n} C_{n - r},$
$\Rightarrow 3 r - 1 + r + 1 = 2 n$
i.e. $4 r = 2 n$ or $n = 2 r$