By the binomial theorem,
[tex]\displaystyle (1+ax)^p = \sum_{i=0}^p \binom pi (ax)^i[/tex]
For i = 1, the corresponding term in the expansion is
[tex]\dbinom p1 ax = apx = 20x[/tex]
so that ap = 20, while for i = 2 we get
[tex]\dbinom p2 (ax)^2 = \dfrac{a^2p(p-1)}2x^2 = 160x^2[/tex]
so that a²p(p - 1)/2 = 160.
Solve for a and p. Observe that
(a²p(p - 1)/2) / (ap) = 160/20
a(p - 1)/2 = 8
ap/2 - a/2 = 8
10 - a/2 = 8
a/2 = 2
a = 4
and it follows that
ap = 20
4p = 20
p = 5
