Let the coefficients of three consecutive terms in the binomial expansion of be in the ratio . Then the coefficient of the term, which is in the middle of these three terms, is _______.
Show answer and solution
Answer: 1120
Solution
Let the three consecutive terms be T(r), T(r+1), T(r+2) with coefficients C(n,r-1)·2^(r-1), C(n,r)·2r, C(n,r+1)·2^(r+1).
From ratio of first to second: C(n,r-1)·2^(r-1) / C(n,r)·2r = 2/5
C(n,r-1)/C(n,r) · (1/2) = 2/5 → r/(n-r+1) · (1/2) = 2/5 → 5r = 4(n-r+1) → 5r = 4n - 4r + 4 → 9r = 4n + 4 ... (1)
From ratio of second to third: C(n,r)·2r / C(n,r+1)·2^(r+1) = 5/8
(r+1)/(n-r) · (1/2) = 5/8 → 4(r+1) = 5(n-r) → 4r + 4 = 5n - 5r → 9r = 5n - 4 ... (2)
From (1) and (2): 4n + 4 = 5n - 4 → n = 8, then 9r = 36 → r = 4.
The middle term is T(r+1) = T(5), with coefficient C(8,4)·24 = 70·16 = 1120.
From ratio of first to second: C(n,r-1)·2^(r-1) / C(n,r)·2r = 2/5
C(n,r-1)/C(n,r) · (1/2) = 2/5 → r/(n-r+1) · (1/2) = 2/5 → 5r = 4(n-r+1) → 5r = 4n - 4r + 4 → 9r = 4n + 4 ... (1)
From ratio of second to third: C(n,r)·2r / C(n,r+1)·2^(r+1) = 5/8
(r+1)/(n-r) · (1/2) = 5/8 → 4(r+1) = 5(n-r) → 4r + 4 = 5n - 5r → 9r = 5n - 4 ... (2)
From (1) and (2): 4n + 4 = 5n - 4 → n = 8, then 9r = 36 → r = 4.
The middle term is T(r+1) = T(5), with coefficient C(8,4)·24 = 70·16 = 1120.
[JEE Main 2023 · 29 Jan · Shift 1]