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If n is positive integer and three consecutive coefficients in the expansion of \small (1+x)^n are in the ratio 6 : 33 : 110, then n =
 

Option: 1

4


Option: 2

6


Option: 3

12


Option: 4

16


Answers (1)

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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

\text { Now }{ }^n C_{r-1}:{ }^n C_r=6: 33                   

\Rightarrow 2 n-13 r+2=0                ....(1)

\\and\ { }^n C_r:{ }^n C_{r+1}=33: 110 \\ \Rightarrow 3 n-13 r-10=0      ...(2)

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+12 x+\frac{12 \cdot 11}{2 \cdot 1} x^2+\frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} x^3+\ldots .\right]

So coefficient of II, III and IV terms are 
\\ 12,6 \times 11,2 \times 11 \times 10 \\ So,\ 12: 6 \times 11: 2 \times 11 \quad \times 10=6: 33: 110.

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