Question

The coefficient of three consecutive terms in the expansion of ${(1+a)}^n$ are in ratio $1:7:21$, then find the value of $n$.

Answer 1

First term =${}^n{C}_r$

Second term =${}^n{C}_{r+1}$

Third term =${}^n{C}_{r+2}$

$\frac{1}{7}=\frac{{}^n{C}_r}{{}^n{C}_{r+1}}=\frac{n!}{r!|n-r|!}=\frac{|r+1|(n-r-1)1}{n!}($

$\frac{1}{7}=\frac{r+1}{n-r}$

$n-r=7r+7$......(1)

$\frac{7}{21}=\frac{1}{3}=\frac{n!}{(r+1)!(n-r-1)!}\times \frac{(r+2)!(n-r-2)}{n!}$

$\frac{1}{3}=\frac{r+2}{n-r-1}$

$n-r-1=3r+6$

$n=4r+7$...........(2)

$n=8r+7$........(1)

$r=0$

$\therefore n=7$