The second term in the expansion of (1-x)(1+2x)Λn is 19x. Find the value of n
(1+2x)^n = 1 + n(2x) + n(n-1)/2! (2x) + n(n-1)(n-2)/3! (2x)^3 + ....
so (1-x)(1+2x)^2
= (1-x)(1 + n(2x) + n(n-1)/2! (2x)^2 + n(n-1)(n-2)/3! (2x)^3 + ....)
= 1 + 2nx + 4n(n-1)/2 x^2 + 8n(n-1)(n-2)/6 x^3 + ... - x - 2nx^2 - 4n(n-1)/2 x^3 - ..
so the only two terms with a first degree x term are
2nx - x
the 2nx - x = 19
2n - 1 = 19
2n = 20
n = 10
To find the value of n, we need to look at the second term in the expansion of (1-x)(1+2x)Λn, which is 19x.
Let's expand (1-x)(1+2x)Λn using the binomial theorem:
(1-x)(1+2x)Λn = Σ (n C r) * (1)Λ(n-r) * (-x)Λr * (1+2x)Λ(r)
where Σ represents the sum over r from 0 to n, C represents the binomial coefficient or "choose" function, and Λ represents the exponentiation operator.
The second term in this expansion will have r = 1, so we can write it as:
(n C 1) * (1)Λ(n-1) * (-x)Λ1 * (1+2x)Λ(1)
Simplifying this expression:
(n) * (1)^(n-1) * (-x) * (1+2x)
Since we are given that the coefficient of x is 19 in the second term, we can equate it to 19x:
(n) * (1)^(n-1) * (-x) * (1+2x) = 19x
Simplifying further:
(n) * (1)^(n-1) * (-1) * (1+2x) = 19
Now, we need to solve this equation to find the value of n.