Last updated at Sept. 24, 2018 by Teachoo
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Prove binomial theorem by mathematical induction. i.e. Prove that by mathematical induction, (a + b)^n = πΆ(π,π) π^(πβπ) π^π for any positive integer n, where C(n,r) = π!(πβπ)!/π!, n > r We need to prove (a + b)n = β_(π=0)^πβγπΆ(π,π) π^(πβπ) π^π γ i.e. (a + b)n = β_(π=0)^πβγππΆππ^(πβπ) π^π γ Let P (n) : (a + b)n = β_(π=0)^πβγππΆππ^(πβπ) π^π γ For n = 1, L.H.S = (a + b)1 = a + b R.H.S = β_(π=0)^1βγ1πΆππ^(1βπ) π^π γ = 1πΆ0π^(1β0) π^0 + 1πΆ1π^(1β1) π^1 = 1!(1 β 0)!/0! π^1 π^0 + 1!(1 β1)!/1! π^0 π^1 = (1 Γ 1)/1 π^1 Γπ^0 + (1 Γ 1)/1 π^0 π^1 = a+b Hence, L.H.S. = R.H.S β΄ P(n) is true for n = 1 Assume that P(k) is true (a + b)k = (a + b)k = kC0 ak b0 + kC1 ak β 1 b1 + β¦. β¦. + kCk β 1 a1 bk β 1 + kCk a0 bk (a + b)k = ak + kC1 ak β 1 b + β¦. β¦. + kCk β 1 a1 bk β 1 + bk We will prove that P(k + 1) is true. (a + b)k + 1 =β_(π=0)^(π+1)βγ πΆππ^(π+1β π) π^π γ (a + b)k = k+1C0 ak+1 b0 + k+1C1 ak + 1 β 1 b1 β¦. β¦. β¦β¦.β¦. + k+1Ck β 1 a1 bk+ 1 β 1 + k+1Ck a0 bk+1 (a + b)k = ak+1 + k+1C1 ak b + β¦. β¦. + k+1Ck β 1 a bk + bk+1 We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) (a + b)k = ak + kC1 ak β 1 b + β¦. β¦. + kCk β 1 a1 bk β 1 + bk Multiplying (a + b) both sides, (a + b)k (a + b) = (a + b) (ak + kC1 ak β 1 b + β¦. β¦. + kCk β 1 a1 bk β 1 + bk ) (a + b)k + 1 = a (ak + kC1 ak β 1 b + β¦. β¦. + kCk β 1 a1 bk β 1 + bk ) + b (ak + kC1 ak β 1 b + β¦. β¦. + kCk β 1 a1 bk β 1 + bk ) (a + b)k + 1 = ak+1 + kC1 ak b + β¦. β¦. + kCk β 1 a2 bk β 1 + abk + bak + kC1 ak β 1 b2 + β¦. β¦. + kCk β 1 abk + bk+1 (a + b)k + 1 = ak+1 + (kC1 ak b + akb) +β¦β¦ β¦β¦. + (kCk β 1 abk + abk) + bk+1 Using nCk + nCk β 1 = n + 1Ck (a + b)k + 1 = ak+1 + k+1C1 ak b +β¦β¦ β¦β¦. + k+1Ck abk + bk+1 which is the same as P(k+1) β΄ P(k+1) is true when P(k) is true β΄ By the principle of mathematical induction, P(n) is true for n, where n is a natural number
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