If three consecutive coefficients in the expansion of $$(x + y)^{n}$$ are in the ratio 1 : 9 : 63, then the value of n is ___________.

The coefficient of any term in the series will be of the form $$^nC_r$$

Let us assume the three consecutive terms to be $$^nC_{r-1},^nC_r,^nC_{r+1}$$

$$\frac{^nC_{r-1}}{^nC_r}=\frac{1}{9}$$

$$\frac{\left(\dfrac{n!}{\left(r-1\right)!\cdot\left(n-r+1\right)!}\right)}{\dfrac{n!}{\left(r\right)!\cdot\left(n-r\right)!}}=\frac{1}{9}$$

$$\left(\frac{\left(r\right)!\cdot\left(n-r\right)!}{\left(r-1\right)!\cdot\left(n-r+1\right)!}\right)=\frac{1}{9}$$

$$\frac{r}{\left(n-r+1\right)}=\frac{1}{9}$$

$$9r=n-r+1\rightarrow 1$$

$$\frac{^nC_{r}}{^nC_{r+!}}=\frac{9}{63}$$

$$\frac{\left(\dfrac{n!}{r!\cdot\left(n-r\right)!}\right)}{\dfrac{n!}{\left(r+1\right)!\cdot\left(n-r-1\right)!}}=\frac{1}{7}$$

$$\left(\frac{\left(r+1\right)!\cdot\left(n-r-1\right)!}{r!\cdot\left(n-r\right)!}\right)=\frac{1}{7}$$

$$\left(\frac{r+1}{\left(n-r\right)}\right)=\frac{1}{7}$$

7r+7=n-r

n= 8r+7$$ \rightarrow 2$$

On solving equations 1 and 2, we get r = 4 and n = 39