Question
If $$'n'$$ is positive integer and three consecutive coefficient in the expansion of $${\left( {1 + x} \right)^n}$$ are in the ratio $$6 : 33 : 110$$ then n is equal to :
A.
9
B.
6
C.
12
D.
16
Answer :
12
Solution :
Let the consecutive coefficient of $${\left( {1 + x} \right)^n}$$ are $$^n{C_{r - 1}},{\,^n}{C_r},{\,^n}{C_{r + 1}}$$
From the given condition, $$^n{C_{r - 1}}:{\,^n}{C_r}:{\,^n}{C_{r + 1}} = 6:33:110$$
Now, $$^n{C_{r - 1}}:{\,^n}{C_r}= 6:33$$
$$\eqalign{
& \Rightarrow \,\frac{{n!}}{{\left( {r - 1} \right)!\left( {n - r + 1} \right)!}} \times \frac{{r!\left( {n - r} \right)!}}{{n!}} = \frac{6}{{33}} \cr
& \Rightarrow \,\frac{r}{{n - r + 1}} = \frac{2}{{11}} \cr
& \Rightarrow \,11r = 2n - 2r + 2 \cr
& \Rightarrow \,2n - 13r + 2 = 0\,\,\,.....\left( {\text{i}} \right) \cr
& {\text{and}}{{\text{ }}^n}{C_r}:{\,^n}{C_{r + 1}} = 33\,:\,110 \cr
& \Rightarrow \,\frac{{n!}}{{r!\left( {n - r} \right)!}} \times \frac{{\left( {r + 1} \right)!\left( {n - r - 1} \right)!}}{{n!}} = \frac{{33}}{{110}} = \frac{3}{{10}} \cr
& \Rightarrow \,\frac{{\left( {r + 1} \right)}}{{n - r}} = \frac{3}{{10}} \cr
& \Rightarrow \,3n - 13r - 10 = 0\,\,\,.....\left( {{\text{ii}}} \right) \cr} $$
Solving (i) & (ii), we get $$n = 12$$