1. 1(1)5 + 5(1)4(2x) + 10(1)3(2x)2 + 10(1)2(2x)3 + 5(1)(2x)4 + 1(2x)5
Binomial theorem
(x+y)1 =
1x + 1y
(x+y)2
1x2 + 2xy + 1y2
(x+y)3
1x3 + 3x2y + 3xy2 + 1y3
(x+y)4
1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x+y)5
1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
(1 + 2x)5 =
1(1)5 + 5(1)4(2x) + 10(1)3(2x)2 + 10(1)2(2x)3 + 5(1)(2x)4 + 1(2x)5
= x + 40x2 + 80x3 + 80x4 + 32x5

2. (x+y)1 =
1x + 1y
(x+y)2
1x2 + 2xy + 1y2
(x+y)3
1x3 + 3x2y + 3xy2 + 1y3
(x+y)4
1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x+y)5
1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
(1 - 2x)5 =
1(1)5 + 5(1)4(-2x) + 10(1)3(-2x)2 + 10(1)2(-2x)3 + 5(1)(-2x)4 + 1(-2x)5
= x + 40x2 - 80x3 + 80x4 - 32x5

3. = 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2)2 + 4(1) (x+x2)3 + 1(x+x2)4
Expand (1 + x + x2)4
Treat as (1 + (x + x2))4
= 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2) (1) (x+x2)3 + 1(x+x2)4
(x+x2)2 = x2 +2x3+x4
(x+x2)3 = x3 +3x2(x2)+3x(x2)2+(x2)3
= x3 + 3x4 + 3x5 + x6
Similarly (x+x2)4 = x4 +4x3(x2) +6x2(x2)2
+4x(x2)3 +(x2)4
= x4 +4x5 +6x6 +4x7 +x8

4. = 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2)2 + 4(1) (x+x2)3 + 1(x+x2)4
Expand (1 + x + x2)4
= (1 + (x + x2))4
= 1(1)4 + 4(1)3(x+x2) + 6(1)2(x+x2) (1) (x+x2)3 + 1(x+x2)4
= 1 + 4x +4x2 +6(x2+2x3+x4)
+4( x3 + 3x4 + 3x5 + x6)
+ x4 +4x5 +6x6 +4x7 +x8
= 1+ 4x + 10x2 + 16x3 + 19x4 + 16x5
+ 10x6 + 4x7 + x8

5. (1 - 2x)5 =
1(1)5 + 5(1)4(-2x) + 10(1)3(-2x)2 + 10(1)2(-2x)3 + 5(1)(-2x)4 + 1(-2x)5
= x + 40x2 - 80x3 + 80x4 - 32x5
Remember nCr
= n!
(n-r)!r!
5C1 = 5
5C2 = 10
5C4 = 5
5C3 = 10
(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

6. Find the coefficient of x5y6 in the expansion of (x+y)11
(2y)6 = 26y6
So coefficient is 462 x 26 = 29568

7. Find the coefficient of x3 in the expansion of (x2+1/x)6
You get x3 by (x2)3 * (1/x)3
So you need 6C3
6C3 = 20
So coefficient is x3 = 20

8. 1(1)5 + 5(1)4(-2x) + 10(1)3(-2x)2 + 10(1)2(-2x)3 + 5(1)(-2x)4 + 1(-2x)5
Find to 3 sf
Let x = 0.01 in the expression
0.985 = 1 – 10(0.01) + 40(0.01)2 – 80(0.01)3 + 80(0.01)4 – 32(0.01)5
0.985 = 1 – – ……
0.985 = =
0.904 (3sf)