Binomial Expansion questions
Slides in this set (7)
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- 1. Expand: i). (z^2-1)^5 ii). (2x-y)^3
- 2. What are the first 3 terms of (2+3x)^10
- 3. Find the coefficient of x^8 in (2-x^2)^9
- 4. The first 3 terms of (1+2x)^n are 1+ax+112x^2. Find a and n
- 5. The first 3 terms of (1+2x)^n are 1+20x+ax^2. Find a and n
- 6. Without using a calculator, find 7C2
- 7. Find n when [n][2] = 45
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- 1. Expand: i). (z^2-1)^5 ii). (2x-y)^3
- i). [5][0]a^5 + [5][1]a^4b + [5][2]a^3b^2 + [5][3]a^2b^3 + [5][4]ab^4 + [5][5]b^5 ;
- [5][0](z^2)^5 + [5][1](z^2)^4 x (-1) + [5][2](z^2)^3 x (-1)^2 + [5][3](z^2)^2 x (-1)^3 + [5][4](z^2) x (-1)^4 + [5][5](-1)^5 =
- z^10-5z^8+16z^6-10z^4+5z^2-1
- ii). [3][0]a^3 + [3][1]a^2b + [3][2]ab^2 + [3][3]b^3 ; [3][0](2x)^3 + [3][1](2x)^2 x (-y) + [3][2](2x) x (-y)^2 + [3][3](-y)^3 =
- 8x^2 - 12x^2y + 6xy^2 -y^3
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- 2. What are the first 3 terms of (2+3x)^10
- [10][0]a^10 + [10][1]a^9b + [10][2]a^8b^2 , 1(2^10) + 10(2^9)(3x) + 45(2^8)(3x^2) ;
- 1024 + 15360x + 103080x^2
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- 3. Find the coefficient of x^8 in (2-x^2)^9
- Inside the brackets we have x^2, and to make x^8 we need to do (x^2)^4 , so we need r=4.
- From here we do [9][4](2^(9-4)) , which is the same as [9][4](2^5). [9][4] = 126, and 126 x 2^5 = 4032.
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- 4. The first 3 terms of (1+2x)^n are 1+ax+112x^2. Find a and n
- [n][0]a^n + [n][1]a^n-1b + [n][2]a^n-2b^2
- Use facts we know like: [n][0] = 1 , [n][1] = n , [n][2] = n(n-1)/2
- [n][0]1^n + [n][1]1^n-1 x 2x + [n][2]1^n-2 x 2x^2 ; 1+n(2x)+n(n-1)/2 x (2x^2) , 1+n(2x)+2n(n-1)x^2
- Equate x^2 : 2n(n-1) = 112 , 2n^2 - 2n - 112 = 0 , n^2 - n - 56 = 0 , (n-8)(n+7) , n=8 as it must be positive
- Equate x : 2n = a , a = 2 x 8 , a=16
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- 5. The first 3 terms of (1+2x)^n are 1+20x+ax^2. Find a and n
- [n][0]a^n + [n][1]a^n-1b + [n][2]a^n-2b^2
- [n][0]1^n + [n][1]1^n-1 x 2x + [n][2]1^n-2 x 2x^2 ; 1+n(2x)+n(n-1)/2 x (2x^2) , 1+n(2x)+2n(n-1)x^2
- 1+n(2x)+n(n-1)/2(2x^2) , 1+2xn=2n(n-1)x^2
- Equate x^2 : 2n(n-1)=a , 2n^2-2n=a. Here we can't go any further, so lets equate x first
- Equate x : 2n=20 , so n=10
- Now to find a we can sub n=10 into the previous equation he has, so 2(10)^2 - 2(10) =a , 200-20 = a, a=180
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- 6. Without using a calculator, find 7C2
- Use the equation : n!/r!(n-r)! , so 1x2x3x4x5x6x7 / 1x2x1x2x3x4x5 = 21
- 7. Find n when [n][2] = 45
- We know [n][2] is n(n-1)/2 , so we can create the equation n(n-1)/2=45 , so n(n-1)=90, n^2-n-90=0 and solve to get (n-10)(n+9). We know n must equal 10 as it must be positive.