Pascals Triangle and the Binomial Theorem x y0

Pascal’s Triangle and the Binomial Theorem (x + y)0 = 1 (x + y)1 = 1 x + 1 y (x + y)2 = 1 x 2 + 2 xy + 1 y 2 (x + y)3 = 1 x 3 + 3 x 2 y + 3 xy 2 +1 y 3 (x + y)4 = 1 x 4 + 4 x 3 y + 6 x 2 y 2 + 4 xy 3 + 1 y 4 (x + y)5 = 1 x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 xy 4 + 1 y 5 (x + y)6 = 1 x 6 + 6 x 5 y 1 + 15 x 4 y 2 + 20 x 3 y 3 + 15 x 2 y 4 + 6 xy 5 + 1 y 6 7. 5. 2

The Binomial Theorem is a formula used for expanding powers of binomials. Each term of the answer is the (a + b)3 = (a + b)(a + b) = a 3 + 3 a 2 b + 3 ab 2 + b 3 product of three first-degree factors. For each term of the answer, 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 3 C 0 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 3 C 1 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 3 C 2 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 3 C 3 is the coefficient of the third term. (a + b)3 = 3 C 0 a 3 + 3 C 1 a 2 b + 3 C 2 ab 2 + 3 C 3 b 3 7. 5. 3

Pascal’s Triangle and the Binomial Theorem The numerical coefficients in a binomial expansion can be found in Pascal’s triangle. n=0 2 nd Row n = 1 3 rd Row n = 2 4 th Row n = 3 1 st Row (a + b)0 (a + b)1 (a + b)2 (a + b)3 n = 4 (a + b)4 5 6 th Row n = 5 (a + b) Pascal’s Triangle Using Combinatorics 1 0 C 0 2 1 3 1 4 1 5 th Row 1 5 10 2 C 0 1 3 6 1 3 C 0 1 4 10 1 C 1 1 C 0 1 1 5 4 C 0 1 5 C 0 2 C 1 3 C 1 4 C 1 5 C 1 3 C 2 4 C 2 5 C 2 2 C 2 3 C 3 4 C 3 5 C 3 4 C 4 5 C 5 7. 5. 4

Binomial Expansion - the General Term (a + b)3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 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 = 3 C 0 a 3 - 0 b 0 + 3 C 1 a 3 - 1 b 1 + 3 C 2 a 3 - 2 b 2 + 3 C 3 a 3 - 3 b 3 (a + b)n = n. C 0 an - 0 b 0 + n. C 1 an - 1 b 1 + n. C 2 an - 2 b 2 + … + n. Ck an - kbk The general term is the (k + 1)th term: tk + 1 = n. Ck an - k bk 7. 5. 5

Binomial Expansion - Practice Expand the following. a) (3 x + 2)4 n=4 a = 3 x b=2 = 4 C 0(3 x)4(2)0 + 4 C 1(3 x)3(2)1 + 4 C 2(3 x)2(2)2 + 4 C 3(3 x)1(2)3 + 4 C 4(3 x)0(2)4 = 1(81 x 4) + 4(27 x 3)(2) + 6(9 x 2)(4) + 4(3 x)(8) + 1(16) = 81 x 4 + 216 x 3 + 216 x 2 + 96 x +16 b) (2 x - 3 y)4 n=4 a = 2 x b = -3 y = 4 C 0(2 x)4(-3 y)0 + 4 C 1(2 x)3(-3 y)1 + 4 C 2(2 x)2(-3 y)2 + 4 C 3(2 x)1(-3 y)3 + 4 C 4(2 x)0(-3 y)4 = 1(16 x 4) + 4(8 x 3)(-3 y) + 6(4 x 2)(9 y 2) + 4(2 x)(-27 y 3) + 81 y 4 = 16 x 4 - 96 x 3 y + 216 x 2 y 2 - 216 xy 3 + 81 y 4 7. 5. 6

Binomial Expansion - Practice c) (2 x - 3 x-1)5 = 5 C 0(2 x)5(-3 x-1)0 + 5 C 1(2 x)4(-3 x-1)1 + 5 C 2(2 x)3(-3 x-1)2 n=5 a = 2 x b = -3 x-1 + 5 C 3(2 x)2(-3 x-1)3 + 5 C 4(2 x)1(-3 x-1)4 + 5 C 5(2 x)0(-3 x-1)5 = 1(32 x 5) + 5(16 x 4)(-3 x-1) + 10(8 x 3)(9 x-2) + 10(4 x 2)(-27 x-3) + 5(2 x)(81 x-4) + 1(-81 x-5) = 32 x 5 - 240 x 3 + 720 x 1 - 1080 x-1 + 810 x-3 - 81 x-5 7. 5. 7

Finding a Particular Term in a Binomial Expansion a) Find the eighth term in the expansion of (3 x - 2)11. tk + 1 = n. Ck an - k bk n = 11 a = 3 x b = -2 k=7 t 7 + 1 = 11 C 7 (3 x)11 - 7 (-2)7 t 8 = 11 C 7 (3 x)4 (-2)7 = 330(81 x 4)(-128) = -3 421 440 x 4 b) Find the middle term of (a 2 - 3 b 3)8. n=8 a = a 2 b = -3 b 3 k=4 tk + 1 = n. Ck an - k bk t 4 + 1 = 8 C 4 (a 2) 8 - 4 (-3 b 3)4 t 5 = 8 C 4 (a 2) 4 (-3 b 3)4 = 70 a 8(81 b 12) = 5670 a 8 b 12 n = 8, therefore, there are nine terms. The fifth term is the middle term. 7. 5. 8

Finding a Particular Term in a Binomial Expansion Find the constant term of the expansion of n = 18 a = 2 x b = -x-2 k=? tk + 1 = n. Ck an - k bk tk + 1 = 18 Ck (2 x)18 - k (-x-2)k tk + 1 = 18 Ck 218 - k x 18 - k (-1)k x-2 k tk + 1 = 18 Ck 218 - k (-1)k x 18 - 3 k = x 0 18 - 3 k = 0 -3 k = -18 k=6 Substitute k = 6: t 6 + 1 = 18 C 6 218 - 6 (-1)6 x 18 - 3(6) t 7 = 18 C 6 212 (-1)6 Therefore, the constant term is 76 038 144. t 7 = 76 038 144 7. 5. 9

Finding a Particular Term in a Binomial Expansion Find the numerical coefficient of the x 11 term of the expansion of n = 10 a = x 2 b = -x-1 k=? tk + 1 = n. Ck an - k bk tk + 1 = 10 Ck (x 2)10 - k (-x-1)k tk + 1 = 10 Ck x 20 - 2 k (-1)k x-k tk + 1 = 10 Ck(-1)k x 20 - 2 k x-k tk + 1 = 10 Ck (-1)k x 20 - 3 k = x 11 20 - 3 k = 11 -3 k = -9 k=3 Substitute k = 3: t 3 + 1 = 10 C 3 (-1)3 x 20 - 3(3) t 4 = 10 C 3 (-1)3 x 11 t 4 = -120 x 11 Therefore, the numerical coefficient of the x 11 term is -120. 7. 5. 10

Finding a Particular Term of the Binomial One term in the expansion of (2 x - m)7 is -15 120 x 4 y 3. Find m. tk + 1 = n. Ck an - k bk n=7 a = 2 x b = -m k=3 tk + 1 = 7 C 3 (2 x)4 (-m)3 tk + 1 = (35) (16 x 4) (-m)3 -15 120 x 4 y 3 = (560 x 4) (-m)3 Therefore, m is 3 y. 3 y = m 7. 5. 11