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 $r =$

\frac{n}{2}, n

even

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\frac{n}{2}

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n

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1

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Solution

The correct option is A $\frac{n}{2}, n$ even
$t_{3 r} = {^{2 n} C}_{3 r - 1} x^{3 r - 1}$ and

$t_{r + 2} = {^{2 n} C}_{r + 1} x^{r + 1}$

Given, ${^{2 n} C}_{r + 1} = {^{2 n} C}_{3 r - 1}$

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

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

$\Rightarrow r = n$ or $r = \frac{1}{2} n$

$r = 1$ is not possible. So $r$ being a positive integer greater than $1$ ,
will be equal to $\frac{1}{2} n$ provided $n$ is an even integer greater than $2$ otherwise $r$ has no value.

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