Question

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

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

Answer 1

The correct option is A

$n=2r$

Explanation for the correct option:

Step$1$. To find ${\left(3r\right)}^{th}$and ${\left(r+2\right)}^{th}$terms:

In the expansion ${(1+x)}^{2n}$

$T_{3r}={}^{2n}C_{3r-1}{\left(x\right)}^{3r-1}$ and

$T_{r+2}={}^{2n}C_{r+1}{\left(x\right)}^{r+1}$

Step2. Equating coefficients of$T_{3r}$ and $T_{r+2}$:

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

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

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

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

But given$r>1$, So $n=2r$

Hence the correct option is A.