(b) We know that the expansion of (1 + x)^n is given by the binomial theorem:
(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n)x^n
We are given that there exist three consecutive coefficients a, b, c such that a:b:c = 1:8:40. Let's denote these coefficients as C(n, k), C(n, k + 1), and C(n, k + 2), where k is some integer. Then we have:
a:b = C(n, k):C(n, k + 1) = 1:8
b:c = C(n, k + 1):C(n, k + 2) = 8:40 = 1:5
Multiplying these two ratios, we get:
a:b:c = 1:8:40
Therefore, we have:
C(n, k) = a
C(n, k + 1) = 8a
C(n, k + 2) = 40a
Using the formula for the binomial coefficients, we can express these coefficients in terms of n and k:
C(n, k) = n! / (k!(n - k)!)
C(n, k + 1) = n! / ((k + 1)!(n - k - 1)!)
C(n, k + 2) = n! / ((k + 2)!(n - k - 2)!)
We can then use these equations to eliminate a and simplify the ratios:
a:b:c = 1:8:40 n! / (k!(n - k)!) : n! / ((k + 1)!(n - k - 1)!) : n! / ((k + 2)!(n - k - 2)!) = 1:8:40
Simplifying this equation, we get:
(k + 2)(k + 1) = 40(n - k)
Expanding the left side and simplifying, we get:
k^2 + 3k - 38n + 40 = 0
We can use the quadratic formula to solve for k:
k = (-3 ± sqrt(9 + 152n)) / 2
Since k is an integer, we need the discriminant to be a perfect square:
9 + 152n = m^2
Solving for n, we get:
n = (m^2 - 9) / 152
Since n is an integer, m must be odd. Trying odd values of m, we find that the smallest value that works is m = 13, which gives:
n = (13^2 - 9) / 152 = 16
