the binomial theorem

Question 1

(1+ax)ⁿ

Answer 1

1+nax+(n(n-1)/2!)(ax)²+(n(n-1)(n-2)3!)(ax)³

Question 2

expand (1-2x)⁴

Answer 2

1+4(-2x)+(4(3)/2!)(-2x)² +(4(3)(2)3!)(ax)³+ (4x3x2x1/4!)(-2x)⁴

1+8x+24x²-32x³+16x⁴

Question 3

(2+6x)⁵

Answer 3

2⁵(1+3x)⁵

(1+3x)⁵= 1+5(3x)+(5(5-1)/2!)(3x)² +(5(5-1)(5-2)3!)(3x)³+(5x4x3x2/4!)(3x)⁴+(5x4x3x2x1/5!)(3x)⁵

1+15x+90x²+270x³+405x⁴+243x⁵

2⁵=32, so times everything by 32

32+480x+2880x²+8640x³+1290x⁴+7776x⁵

Question 4

(1+x)^-2

Answer 4

1+(-2)(x)+((-2)(-3)/2!)(x²)+((-2)(-3)(-4)/3!)(x³)+((-2)(-3)(-4)(-5)/4!)(x⁴)

=1-2x+3x²-4x³+5x⁴

Question 5

words of wisdom

Answer 5

don’t let your heart speak louder than your mind

Question 6

how to prove |3x/4|<1 is equivalent to |x|<4/3

Answer 6

• 1<3x/4<1
• 4/3<x></x>

<p>|x|&lt;4/3</p>

</x>

Question 7

binomial expansion of 1/(1+3x)² to x³

and use it to find an approximate v value of sqrt(0.94)

Answer 7

1+(-2)(3x)+((-2)(-3)/(2!))(-2x)²+((1/2)(-1/2)(-3/2)/3!)(-2x)³

=1-x-1/2x²-x³/2

-1<-2x<1

valid when |-2x|<1 which gives x<1/2

sqrt(0.94)=0.94^1/2=(1-2x)^1/2 so x=0.03

|0.03|<1/2 so the expansion is valid for this value

sqrrt(0.94) =(1-2x0.03)^1/2

=1-(0.03)-1/2(0.03)-1/2(0.03)² -1/2(0.03)³

=0.97

Question 8

(3+x)^-1up to x³ find what it is valid for

Answer 8

(3+x)^-1= 3^-1(1+x/3)^-1=1/3(1+x/3)^-1

=1/3(1+(-1)(x/3) + ((-1)(-2)/2!)(x/3)²+((-1)(-2)(-3)/3!)(x/3)³

=1/3 - (1/9)x + (1/27)x² - (1/81)x³

valid when |(1/3)x|<1 so |x|<3

show |-4x|<1 is equivalent to |x|<1/4

-1<-4x<1, 1>4x>-1, -14x<1, -1/4<x>
</x>

Question 9

(1+x)^2/3 to x² use to find value of 1.6^2/3 how accurate?

Answer 9

(1+x)^2/3=1+(2/3)x +((2/3)(-1/3)/2!)(x)²=1+(2/3)x-(1/9)x²

1.6^2/3=(1+0.6)^2/3 = 1+(2/3)(0.6) - 1/9(0.6)²

|0.6|<1 so we can use the expansion

=1+0.4-0.04 = 1.36

1.6^2/3=1.36798, accurate to two significant figures (1.4)

Question 10

(1+3x)/(1+x)³

|x|<1

to x³

Answer 10

(1+3x)(1+x)^-3

(1+3x)(1 + (-3)x + (-3)(-4)(x²)/2! + (-3)(-4)(-5)(x³)/3! + …)

(1+3x)(1-3x+6x²-10x³+…)

=1-3x+6x²-10x³+…+3x-9x²+18x³-…

1-3x²+8x³

Question 11

f(x)= (10-x)/(3-x)(1+2x)

expand as far as x², state the range

Answer 11

A/(3-x) + B/(1+2x)

1/(3-x) + 3/(1+2x)

1/(3-x)=(3-x)^-1=3^-1(1-x/3)^-1=1/3(1-x/3)^-1

expanded: 1/3 + (1/9)x + (1/27)x²+…

3/(1+2x)= 3(1+2x)^-1=3(expansion)= 3-6x+12x²-…

add together: 10/3 -(53/9)x + (325/27)x²

expansion for (1-x/3)^-1 valid for |-1/3|<1 gives |x|< 3

expansion for (1-2x)^-1 valid for |2x|<1 gives |x|< 1/2

so its the one that satisfies both inequalities: |x|<1/2

Question 12

what must you remember for all expansions

Answer 12

to have +…. or -…. at the end

Question 13

what’s the expansion always valid for

Answer 13

|ax| < 1