Question

If the coefficient of $r^{th}$ and $(r+1)th$ term in the expansion of $(3+7x{)}^{29}$ are equal, then $r$ equals:

• A. $20$
• B. $21$
• C. $14$
• D. $Noneofthese$

Answer 1

The correct option is A $20$

A] For $(a+bx{)}^n$

$t_r{=}^n{C}_r{a}^{n-r}(bx{)}^r$

Here a = 3 , b = 7 , n = 29 as $(3x+7x{)}^{29}$

$t_r{=}^{29}{C}_r\left(3{)}^{29-r}\right(7x{)}^r$

$t_r+1{=}^{29}{C}_{r+1}\left(3{)}^{29-r-1}\right(7x{)}^{r+1}$

$t_r=t_r+1$

$\Rightarrow \frac{29!}{r!(29-r)!}\times \left(3{)}^{29-r}\right(7x{)}^r=\frac{29!}{(r+1)!}\times \frac{(3{)}^{29-r}}{(29-(r+1\left)\right)}\times \frac{7}{3}$

$\Rightarrow \frac{7^r}{ri(29-r)}=\frac{1}{(r+1)r!}\times \frac{7^r}{(28-r)!}\times \frac{7}{3}$

$\Rightarrow \frac{3}{29-r}=\frac{7}{r+1}$

$\Rightarrow 3r+3=7\times 29-7r$

$\Rightarrow 3r+3=203-7r$

$\Rightarrow 10r=200$

$\therefore r=20$