Proving binomial theorem by mathematical induction - Expansion

Proving P(n) true for n 1 -2.PNG

Taking P(k) true, proving P(k+1) true - 3.PNG

Proving P(k+1) true - 4.PNG

P(n) is true - 5.PNG

  1. Chapter 4 Class 11 Mathematical Induction
  2. Concept wise
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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.