Question

Can someone please help me with this?

Answer 1

By the binomial theorem,

$\displaystyle (1+ax)^p = \sum_{i=0}^p \binom pi (ax)^i$

For i = 1, the corresponding term in the expansion is

$\dbinom p1 ax = apx = 20x$

so that ap = 20, while for i = 2 we get

$\dbinom p2 (ax)^2 = \dfrac{a^2p(p-1)}2x^2 = 160x^2$

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