pascal’s triangle and the binomial theorem (x + y) 0 = 1 (x + y) 1 = 1x + 1y (x + y) 2 = 1x 2 +...
TRANSCRIPT
The Binomial Theorem
Pascal’s Triangle and the Binomial Theorem
(x + y)0 = 1
(x + y)1 = 1x + 1y
(x + y)2 = 1x2 + 2xy + 1y2
(x + y)3 = 1x3 + 3x2y + 3xy2 +1 y3
(x + y)4 = 1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
(x + y)5 = 1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y5
(x + y)6 = 1x6 + 6x5y1 + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
The Binomial Theorem is a formula used for expanding powers of binomials.
(a + b)3 = (a + b)(a + b)(a + b) = a3 + 3a2b + 3ab2 + b3
Each term of the answer is the product of three first-degree factors. For each term of theanswer, an a and/or b is taken from each first-degree factor.
• The first term has no b. It is like choosing no b from three b’s. The combination 3C0 is the coefficient of the first term
• The second term has one b. It is like choosing one b from three b’s. The combination 3C1 is the coefficient of the second term
• The third term has two b’s. It is like choosing two b’s from three b’s. The combination 3C2 is the coefficient of the first term
• The fourth term has three b’s. It is like choosing three b’s from three b’s. The combination 3C3 is the coefficient of the third term
(a + b)3 = 3C0 a3 + 3C1 a2b + 3C2 ab2 + 3C3 b3
The Binomial Theorem
1st Row
2nd Row
3rd Row
4th Row
5th Row
0th Row
The numerical coefficients in a binomial expansion can be found in Pascal’s triangle.
(a + b)n
0C0
1C01C1
2C02C1
2C2
3C03C1
3C23C3
4C04C1
4C24C3
4C4
5C05C1
5C25C3
5C45C5
Pascal’s Triangle and the Binomial Theorem
1
Pascal’sTriangle
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Pascal’s TriangleUsing Combinatorics
(a + b)3 = a3 + 3a2b + 3ab2 + b3
The degree of each term is 3.For the variable a, the degree descends from 3 to 0.For the variable b, the degree ascends from 0 to 3.
(a + b)3 = 3C0 a3 - 0b0 + 3C1 a3 - 1b1 + 3C2 a3 - 2b2 + 3C3 a3 - 3b3
(a + b)n = nC0 an - 0b0 + nC1 an - 1b1 + nC2 an - 2b2 + … + nCk an - kbk
The general term is the (k + 1)th term:
tk + 1 = nCk an - k bk
Binomial Expansion - the General Term
Expand the following.
a) (3x + 2)4 = 4C0(3x)4(2)0 + 4C1(3x)3(2)1
+ 4C2(3x)2(2)2 + 4C3(3x)1(2)3 + 4C4(3x)0(2)4
n = 4a = 3x b = 2
= 1(81x4) + 4(27x3)(2) + 6(9x2)(4)+ 4(3x)(8) + 1(16)
= 81x4 + 216x3 + 216x2 + 96x +16
Binomial Expansion - Practice
Expand the following.
b) (2x - 3y)4
= 4C0(2x)4(-3y)0
+ 4C3(2x)1(-3y)3
= 1(16x4)
= 16x4 - 96x3y + 216x2y2 - 216xy3 + 81y4
n = 4a = 2x b = -3y
Binomial Expansion - Practice
+ 4C1(2x)3(-3y)1 + 4C2(2x)2(-3y)2
+ 4C4(2x)0(-3y)4
+ 4(8x3)(-3y) + 6(4x2)(9y2) + 4(2x)(-27y3) + 81y4
Binomial Expansion - Practice
2x
3
x
5
(2x - 3x-1)5
= 5C0(2x)5(-3x-1)0 + 5C1(2x)4(-3x-1)1 + 5C2(2x)3(-3x-1)2
+ 5C3(2x)2(-3x-1)3 + 5C4(2x)1(-3x-1)4 + 5C5(2x)0(-3x-1)5
n = 5a = 2x b = -3x-1
= 1(32x5) + 5(16x4)(-3x-1) + 10(8x3)(9x-2) + 10(4x2)(-27x-3)+ 5(2x)(81x-4) + 1(-81x-5)
= 32x5 - 240x3 + 720x1 - 1080x-1 + 810x-3 - 81x-5
c)
a) Find the eighth term in the expansion of (3x - 2)11.
tk + 1 = nCk an - k bk
n = 11a = 3xb = -2k = 7t7 + 1 = 11C7 (3x)11 - 7
(-2)7
t8 = 11C7 (3x)4 (-2)7
= 330(81x4)(-128) = -3 421 440 x4
b) Find the middle term of (a2 - 3b3)8.
n = 8a = a2
b = -3b3
k = 4
n = 8, therefore, there are nine terms. The fifth term is the middle term.tk + 1 = nCk an - k
bk
t4 + 1 = 8C4 (a2) 8 - 4 (-3b3)4
t5 = 8C4 (a2) 4 (-3b3)4
= 70a8(81b12) = 5670a8b12
Finding a Particular Term in a Binomial Expansion
Finding a Particular Term in a Binomial Expansion
2x 1
x2
18
.Find the constant term of the expansion of
tk + 1 = nCk an - k bk
n = 18a = 2xb = -x-2
k = ?tk + 1 = 18Ck (2x)18 - k
(-x-2)k
tk + 1 = 18Ck 218 - k x18 - k (-1)k x-2k
tk + 1 = 18Ck 218 - k (-1)k x18 - k x-2k
tk + 1 = 18Ck 218 - k (-1)k x18 - 3k
x18 - 3k = x0
18 - 3k = 0 -3k = -18 k = 6
t6 + 1 = 18C6 218 - 6 (-1)6 x18 - 3(6)
Substitute k = 6:
t7 = 18C6 212 (-1)6
t7 = 76 038 144Therefore, the constantterm is 76 038 144.
Finding a Particular Term in a Binomial Expansion
x2 1
x
10
.
Find the numerical coefficient of the x11 term of the expansion of
tk + 1 = nCk an - k bk
n = 10a = x2
b = -x-1
k = ? tk + 1 = 10Ck (x2)10 - k (-x-1)k
tk + 1 = 10Ck x20 - 2k (-1)k x-k
tk + 1 = 10Ck(-1)k x20 - 2k x-k
tk + 1 = 10Ck (-1)k x20 - 3k
x20 - 3k = x11
20 - 3k = 11 -3k = -9 k = 3
t3 + 1 = 10C3 (-1)3 x20 - 3(3)
Substitute k = 3:
t4 = 10C3 (-1)3x11
t4 = -120 x11
Therefore, the numericalcoefficient of the x11 term is -120.
One term in the expansion of (2x - m)7 is -15 120x4y3. Find m.
Finding a Particular Term of the Binomial
tk + 1 = nCk an - k bkn = 7
a = 2xb = -mk = 3
tk + 1 = 7C3 (2x)4 (-m)3
tk + 1 = (35) (16x4) (-m)3
-15 120x4y3 = (560x4) (-m)3
15 120x4 y3
560x4 m3
27y3 m3
27 y3 m3
27 y33 m
3y = m
Therefore, m is 3y.
The Binomial Theorem
I have made this one [letter] longer than usual because I did not have the leisure to make it shorter.
— Blaise Pascal
