Expand using binomial theorem: (1 + 2x)5.
Using binomial theorem, we have
(1 + 2x)5 = 5C0(1)5 + 5C1(1)4(2x) + 5C2(1)3(2x)2 + 5C3(1)2(2x)3 + 5C4(1)(2x)4 + 5C5(2x)5
= 5!/0!5! × 1 + 5!/1!4! × (2x) + 5!/2!3! × (2x)2 + 5!/3!2! × (2x)3 + 5!/4!1! × (2x)4 + 5!/5!0! × (2x)5
= 1 + 5(2x) + 10(2x)2 + 10(2x)3 + 5(2x)4 + (2x)5
= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5