Question

Given positive integers $r >1, n > 2$ and the coefficient of $(3r)th$ and $(r + 2)th$ terms in the binomial expansion of $(1 + x)^2n$ are equal. Then,

• A. n=2r
• B. n=2r+1
• C. n=3r
• D. None of these

Answer 1

Answer: (A)

n=2r

Answer 2

In the expansion $(1+x)^{2n},t_{3r} = \, ^{2n}C_{3r-1} (x)^{3r-1}$
and \hspace20mm \, t_{r+2}= \, ^{2n}C_{r+1} (x) ^{r+1}
Since, binomial coefficients of t$_3r$ and t$_{r+ 2}$ are equal
$\Rightarrow \, \, \, \, \, \, \, \, 3r-1=r+1 \, or \, 2n=(3r-1)+(r+1)$
$\Rightarrow \, \, \, \, \, \, \, \,$ 2r=2 $\, or \, \, \, \, \, \, \, \,$ 2n = 4r
$\Rightarrow \, \, \, \, \, \, \, \,$ r = 1 $\, or \, \, \, \, \, \, \, \,$ n=2r
But $\, \, \, \, \, \, \, \,$ r >1
$\therefore \, \,$We take, n=2r