Question

If the coefficients of $(3r{)}^{\text{th}}$ and $(r+2{)}^{\text{th}}$ terms in the expansion of $(1+x{)}^{2n}$ are equal $(r>1,n>2)$, then

• A. $n=r$
• B. $n=r+1$
• C. $n=2r$
• D. $n=2r-1$

Answer 1

The correct option is C

$n=2r$

Explanation for the correct option.

Given, $(1+x{)}^{2n}$

The general term of the expansion is given by:

${\mathrm{T}}_{\mathrm{r}+1}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}$
Therefore the coefficient of $3{\mathrm{r}}^{\text{th}}$ term is
${}^2{}^{n}C_{3r-1}\dots \text{(i)}$
Therefore the coefficient of $(\mathrm{r}+2{)}^{\text{th}}$ term is
${}^{2\mathrm{n}}\mathrm{C}_{\mathrm{r}+1}\dots \text{(ii)}$
Equate (1) & (2) according to question:
${}^{2n}C_{3r-1}={}^{2n}C_{r+1}$
Using property ${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}={}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}-\mathrm{r}}$
$\begin{array}{rcl}3\mathrm{r}-1& =& 2\mathrm{n}-\mathrm{r}-1\\ 4\mathrm{r}& =& 2\mathrm{n}\\ \mathrm{n}& =& 2\mathrm{r}\end{array}$
Hence option C is correct.