Expand the binomial expression ( 2x+ 1)5

Using the Binomial theorem,

( 2x+ 1)5 = 5C0(2x)5 10 + 5C1(2x)5-1 11 + 5C2(2x)5-2 12 + 5C3(2x)5-3 13 + 5C4(2x)5-4 14 + 5C5(2x)5-5 15

= 5C0(2x)5 + 5C1(2x)4 + 5C2(2x)3 + 5C3(2x)2 + 5C4(2x)1 + 5C5(2x)0

= 5C032x5 + 5C116x4 + 5C28x3 + 5C34x2 + 5C4 2x1 + 5C5

= 32x5 + (5) 16x4+ (10)8x3 + (10) 4x2 + (5) 2x + 1

= 32x5 + 80x4+ 80x3 + 40x2 + 10x + 1