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Proving binomial theorem by mathematical induction - Expansion

Proving binomial theorem by mathematical induction - Part 2

Proving binomial theorem by mathematical induction - Part 3

Proving binomial theorem by mathematical induction - Part 4

Proving binomial theorem by mathematical induction - Part 5


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Prove binomial theorem by mathematical induction. i.e. Prove that by mathematical induction, (a + b)^n = 𝐢(𝑛,π‘Ÿ) π‘Ž^(π‘›βˆ’π‘Ÿ) 𝑏^π‘Ÿ for any positive integer n, where C(n,r) = 𝑛!(π‘›βˆ’π‘Ÿ)!/π‘Ÿ!, n > r We need to prove (a + b)n = βˆ‘_(π‘Ÿ=0)^𝑛▒〖𝐢(𝑛,π‘Ÿ) π‘Ž^(π‘›βˆ’π‘Ÿ) 𝑏^π‘Ÿ γ€— i.e. (a + b)n = βˆ‘_(π‘Ÿ=0)^π‘›β–’γ€–π‘›πΆπ‘Ÿπ‘Ž^(π‘›βˆ’π‘Ÿ) 𝑏^π‘Ÿ γ€— Let P (n) : (a + b)n = βˆ‘_(π‘Ÿ=0)^π‘›β–’γ€–π‘›πΆπ‘Ÿπ‘Ž^(π‘›βˆ’π‘Ÿ) 𝑏^π‘Ÿ γ€— For n = 1, L.H.S = (a + b)1 = a + b R.H.S = βˆ‘_(π‘Ÿ=0)^1β–’γ€–1πΆπ‘Ÿπ‘Ž^(1βˆ’π‘Ÿ) 𝑏^π‘Ÿ γ€— = 1𝐢0π‘Ž^(1βˆ’0) 𝑏^0 + 1𝐢1π‘Ž^(1βˆ’1) 𝑏^1 = 1!(1 βˆ’ 0)!/0! π‘Ž^1 𝑏^0 + 1!(1 βˆ’1)!/1! π‘Ž^0 𝑏^1 = (1 Γ— 1)/1 π‘Ž^1 ×𝑏^0 + (1 Γ— 1)/1 π‘Ž^0 𝑏^1 = a+b Hence, L.H.S. = R.H.S ∴ P(n) is true for n = 1 Assume that P(k) is true (a + b)k = (a + b)k = kC0 ak b0 + kC1 ak – 1 b1 + …. …. + kCk – 1 a1 bk – 1 + kCk a0 bk (a + b)k = ak + kC1 ak – 1 b + …. …. + kCk – 1 a1 bk – 1 + bk We will prove that P(k + 1) is true. (a + b)k + 1 =βˆ‘_(π‘Ÿ=0)^(π‘˜+1)β–’γ€– πΆπ‘Ÿπ‘Ž^(π‘˜+1βˆ’ π‘Ÿ) 𝑏^π‘Ÿ γ€— (a + b)k = k+1C0 ak+1 b0 + k+1C1 ak + 1 – 1 b1 …. …. …….…. + k+1Ck – 1 a1 bk+ 1 – 1 + k+1Ck a0 bk+1 (a + b)k = ak+1 + k+1C1 ak b + …. …. + k+1Ck – 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 – 1 b + …. …. + kCk – 1 a1 bk – 1 + bk Multiplying (a + b) both sides, (a + b)k (a + b) = (a + b) (ak + kC1 ak – 1 b + …. …. + kCk – 1 a1 bk – 1 + bk ) (a + b)k + 1 = a (ak + kC1 ak – 1 b + …. …. + kCk – 1 a1 bk – 1 + bk ) + b (ak + kC1 ak – 1 b + …. …. + kCk – 1 a1 bk – 1 + bk ) (a + b)k + 1 = ak+1 + kC1 ak b + …. …. + kCk – 1 a2 bk – 1 + abk + bak + kC1 ak – 1 b2 + …. …. + kCk – 1 abk + bk+1 (a + b)k + 1 = ak+1 + (kC1 ak b + akb) +…… ……. + (kCk – 1 abk + abk) + bk+1 Using nCk + nCk – 1 = n + 1Ck (a + b)k + 1 = ak+1 + k+1C1 ak b +…… ……. + k+1Ck abk + bk+1 which is the same as P(k+1) ∴ P(k+1) is true when P(k) is true ∴ By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.