Question
The second term in the expansion of (1-x)(1+2x)Λn is 19x. Find the value of n
Answers
Reiny
(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
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