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Question

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


A

4

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B

6

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C

12

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D

16

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Solution

The correct option is C

12


Let the consecutive coefficient of (1+x)n are

nCr1, nCr, nCr+1

By condition, nCr1 : nCr : nCr+1 = 6 : 33 : 110

Now nCr1 : nCr = 6 : 33

2n - 13r + 2 = 0 ..............(i)

and nCr : nCr+1 = 33: 110

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. Similarity (b).

So alternate (c), n = 12 gives

(1+x)12 =[1+12x+12.112.1 x2 + 12.11.103.2.1x3+.........]

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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