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 =
Answer: Option C
:
C
Let the consecutive coefficient of (1+x)n are
nCr−1,nCr,nCr+1
By condition,nCr−1 :nCr :nCr+1 = 6 : 33 : 110
NownCr−1 :nCr = 6 : 33
⇒2n - 13r + 2 = 0 ..............(i)
andnCr :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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:
C
Let the consecutive coefficient of (1+x)n are
nCr−1,nCr,nCr+1
By condition,nCr−1 :nCr :nCr+1 = 6 : 33 : 110
NownCr−1 :nCr = 6 : 33
⇒2n - 13r + 2 = 0 ..............(i)
andnCr :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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