Maths
posted by Natasha .
The second term in the expansion of (1x)(1+2x)Λn is 19x. Find the value of n

(1+2x)^n = 1 + n(2x) + n(n1)/2! (2x) + n(n1)(n2)/3! (2x)^3 + ....
so (1x)(1+2x)^2
= (1x)(1 + n(2x) + n(n1)/2! (2x)^2 + n(n1)(n2)/3! (2x)^3 + ....)
= 1 + 2nx + 4n(n1)/2 x^2 + 8n(n1)(n2)/6 x^3 + ...  x  2nx^2  4n(n1)/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