A) 4
B) 6
C) 12
D) 16
Correct Answer: C
Solution :
Let the consecutive coefficient of (1+x)n are \[^{n}{{C}_{r-1}},{{\,}^{n}}{{C}_{r}},{{\,}^{n}}{{C}_{r+1}}\] By condition, \[^{n}{{C}_{r-1}}:{{\,}^{n}}{{C}_{r}}{{:}^{n}}{{C}_{r+1}}=6:33:110\] Now\[^{n}{{C}_{r-1}}:{{\,}^{n}}{{C}_{r}}=6:33\] \[\Rightarrow \,2n-13r+2=0\] ?..(i) and \[^{n}{{C}_{r}}:{{\,}^{n}}{{C}_{r+1}}=33:110\] \[\Rightarrow \,\,3n-13r-10=0\] ?..(ii) Solving (i) and (ii), we get n=12 and r=2. Aliter: We take first n=4 [By alternate (a)] but (a) does not hold. Similarly (b). So alternate (c), n =12 gives \[{{(1+x)}^{12}}=\]\[\left[ 1+12x+\frac{12.11}{2.1}{{x}^{2}}+\frac{12.11.10}{3.2.1}{{x}^{3}}+.... \right]\] So coefficient of II, III and IV terms are 12, 6 × 11, 2 × 11 × 10. So, 12 : 6 × 11 : 2 × 11 × 10 = 6 : 33 :110.
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