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 $(1 + x)^{2 n}$ are equal, then

n = 2 r

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n = 2 r + 1

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n = 3 r

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None of these

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Solution

The correct option is A $n = 2 r$
$3 r t h$ term in the expansion of $(1 + x)^{2 n}$

$=^{2 n} C_{3 r - 1} x^{3 r - 1}$

and $(r + 2) t h$ term in the expansion of $(1 + x)$

$=^{2 n} C_{r + 1} x^{r + 1}$

Given that the binomial coefficients of $(3 r - 1)$ and $(r + 2) t h$ terms are equal.

Thus $^{2 n} C_{3 r - 1} =^{2 n} C_{r + 1}$

$\Rightarrow 3 r - 1 = r + 1$

or $2 n = (3 r - 1) + (r + 1)$

$\Rightarrow 2 r = 2$ or $2 n = 4 r$

$\Rightarrow r = 1$ or $n = 2 r$

But $r > 1$

Therefore, $n = 2 r$

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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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Q. Given positive integers

r > 1, n > 2

and the coefficients of

(3 r) t h

term and

(r + 2) t h

term in the binomial expansion of

(1 + x)^{2 n}

are equal then

r =

If for positive integers r > 1 n > 1 and the coefficient of $(3 r)^{t h}$ and $(r + 2)^{t h}$ terms in the binomial expansion of $(1 + x)^{2 n}$ are equal, then