Question

Given the integers $r>1,n>2$ and coeff of $(3r{)}^{th}$ and $(r+2{)}^{nd}$ terms in the binomial expansion of $(1+x{)}^{2n}$ are equal, then

• A. $n=2r$
• B. $n=4r$
• C. $n=2r+1$
• D. None of these

Answer 1

Answer: (B)

$n=4r$

Solution:

We know that in the coefficient of $r^{th}$ term in expansion of $(1+x{)}^n={}^n{C}_{r-1}$

$\therefore$ coefficients of $(3r{)}^{th}$ and $(r+2{)}^{th}$ in the expansion $(1+x{)}^{2n}$ are ${}^{2n}{C}_{3r-1}$ and ${}^{2n}{C}_{r+1}.$

Now,

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

or, $3r-1=r+1$ or, $3r-1+r+1=n$ $(\because {}^n{C}_r={}^{n}Cs$ then $r=s$ or $r+s=n)$

On solving,

or $3r-1=r+1$

or, $r=1$ and

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

or, $4r=n$

or, $n=4r$

Hence, D is the correct option.