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The Binomial Theorem In these notes we prove the binomial theorem, which says that for any integer n ≥ 1, n (x + y) = n X n ℓ X xℓ y n−ℓ = ℓ=0 ℓ+m ℓ ℓ m xy n ℓ where = n! ℓ!(n−ℓ)! (Bn ) ℓ,m≥0 ℓ+m=n Here n! (read “n factorial”) means 1 × 2 × 3 × · · · × n so that, for example, n 1 = n n−1 = n 2 = n n−2 = n! 1!(n−1)! n! 2!(n−2)! = 1×2×3×···×(n−1)×n 1×2×3×···×(n−1) = 1×2×3×···×(n−2)×(n−1)×n (1×2)×(1×2×3×···×(n−2)) = = n n(n−1) 2 n! = 1. As special cases of the Binomial By convention 0! = 1 so that n0 = nn = 0!n! Theorem, we have n = 1 (x + y)1 = x + y n = 2 (x + y)2 = x2 + 2xy + y 2 n = 3 (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3 The proof is by induction on n. First we check Proof of the Binomial Theorem: that, when n = 1, n X ℓ n−ℓ n! ℓ!(n−ℓ)! x y = ℓ n−ℓ n! n=1 ℓ!(n−ℓ)! x y ℓ n−ℓ n! n=1 ℓ!(n−ℓ)! x y + ℓ=0 ℓ=0 = 1! 0 1 0!1! x y + 1! 1 0 1!0! x y ℓ=1 =x+y so that (Bn ) is correct for n = 1. To complete the proof we have to show that, for any integer n ≥ 2, (Bn ) is a consequence of (Bn−1 ). So pick any integer n ≥ 2 and assume that (x + y)n−1 = n−1 X n−1 ℓ ℓ n−1−ℓ xy ℓ=0 is true. Now compute n (x + y) = (x + y) n−1 (x + y) = n−1 X n−1 ℓ x ℓ+1 n−1−ℓ y ℓ=0 c Joel Feldman. 2012. All rights reserved. September 18, 2012 + n−1 X n−1 ℓ xℓ y n−ℓ ℓ=0 The Binomial Theorem 1 The second sum has the same powers of x and y, namely xℓ y n−ℓ , as appear in (Bn ). The make the powers of x and y in the first sum, namely xℓ+1 y n−1−ℓ look more like those of (Bn ), we make the change of summation variable from ℓ to ℓ̃ = ℓ + 1. The first sum n−1 X n−1 ℓ n ℓ+1 n−1−ℓ X x y = ℓ=0 n−1 ℓ̃ n−ℓ̃ xy ℓ̃−1 ℓ̃=1 As ℓ̃ is just a dummy summation variable, we may call it anything we like. In particular, we may rename ℓ̃ back to ℓ. So we now have (x + y)n = n X n−1 ℓ−1 X ℓ n−ℓ n−1 xy + ℓ=1 n−1 ℓ xℓ y n−ℓ ℓ=0 n−1 ℓ n−ℓ = ℓ−1 x y ℓ=n + n−1 ℓ xℓ y n−ℓ ℓ=0 + n−1 X n−1 ℓ−1 + n−1 ℓ xℓ y n−ℓ ℓ=1 Recalling that n! = n (n − 1)! we have n ℓ = n! ℓ!(n−ℓ)! = n(n−1)! ℓ(ℓ−1)! (n−ℓ)! = n n−1 ℓ ℓ−1 n ℓ = n! ℓ!(n−ℓ)! = n(n−1)! ℓ! (n−ℓ)(n−ℓ−1)! = n−1 n n−ℓ ℓ So (x + y)n = n−1 n−1 xn + n−1 0 yn + n−1 X n−1 ℓ−1 + n−1 ℓ ℓ n−ℓ xy ℓ=1 = xn + y n + n−1 X n ℓ ℓ n + n−ℓ n ℓ n−ℓ xy ℓ=1 = xn + y n + n−1 X n ℓ xℓ y n−ℓ ℓ=1 = n X n ℓ xℓ y n−ℓ ℓ=0 as desired. c Joel Feldman. 2012. All rights reserved. September 18, 2012 The Binomial Theorem 2