A) 9
B) 6
C) 12
D) 16
Correct Answer: C
Solution :
[c] Let the consecutive coefficient of |
\[{{(1+x)}^{n}}\] 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 \[^{n}{{C}_{r-1}}:{{\,}^{n}}{{C}_{r}}=6:33\] |
\[\Rightarrow \frac{n!}{(r-1)!(n-r+1)!}\times \frac{r!(n-r)!}{n!}=\frac{6}{33}\] |
\[\Rightarrow \frac{r}{n-r+1}=\frac{2}{11}\Rightarrow 11r=2n-2r+2\] |
\[\Rightarrow 2n-13r+2=0...(i)\] |
and \[^{n}{{C}_{r}}:{{\,}^{n}}{{C}_{r+1}}=33:110\] |
\[\Rightarrow \frac{n!}{r!(n-r)!}\times \frac{(r+1)!(n-r-1)}{n!}=\frac{33}{110}=\frac{3}{10}\] |
\[\Rightarrow \frac{(r+1)}{n-r}=\frac{3}{10}\Rightarrow 3n-13r-10=0..(ii)\] |
Solving (i) & (ii), we get \[n=12\] |
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