Ex 4.1, 12 - Prove a + ar + ar2 + ... + a rn-1 = a(rn - 1)/r-1 - Ex 4.1

Ex 4.1, 12 - Chapter 4 Class 11 Mathematical Induction - Part 2
Ex 4.1, 12 - Chapter 4 Class 11 Mathematical Induction - Part 3 Ex 4.1, 12 - Chapter 4 Class 11 Mathematical Induction - Part 4

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Question 12: Prove the following by using the principle of mathematical induction for all n ∈ N: a + ar + ar2 + ……..+ arn – 1 = (š‘Ž(š‘Ÿ^š‘› āˆ’ 1))/(š‘Ÿ āˆ’ 1) Let P (n) : a + ar + ar2 + ……..+ arn – 1 = š‘Ž(š‘Ÿ^š‘› āˆ’ 1)/(š‘Ÿ āˆ’ 1) For n = 1, L.H.S = a R.H.S = (š‘Ž(š‘Ÿ1 āˆ’ 1))/(š‘Ÿ āˆ’ 1) = (š‘Ž(š‘Ÿ āˆ’ 1))/(š‘Ÿ āˆ’ 1) = a L.H.S. = R.H.S ∓ P(n) is true for n = 1 Assume that P(k) is true a + ar + ar2 + ……..+ ark – 1 = š‘Ž(š‘Ÿ^š‘˜ āˆ’ 1)/(š‘Ÿ āˆ’ 1) We will prove that P(k + 1) is true. a + ar + ar2 + ……..+ ar(k + 1) – 1 = š‘Ž(š‘Ÿ^(š‘˜ + 1) āˆ’ 1)/(š‘Ÿ āˆ’ 1) a + ar + ar2 + ……..+ ark – 1 + ark = š‘Ž(š‘Ÿ^(š‘˜ + 1) āˆ’ 1)/(š‘Ÿ āˆ’ 1) We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) a + ar + ar2 + ……..+ ark – 1 = š‘Ž(š‘Ÿ^š‘˜ āˆ’ 1)/(š‘Ÿ āˆ’ 1) Adding ark both sides a + ar + ar2 + …….. +ark – 1 + ark = š‘Ž(š‘Ÿ^š‘˜ āˆ’ 1)/(š‘Ÿ āˆ’ 1) + ark = (š‘Ž(š‘Ÿ^š‘˜ āˆ’ 1) + (š‘Ÿ āˆ’ 1)š‘Žš‘Ÿ^š‘˜)/(š‘Ÿ āˆ’ 1) = (š‘Žš‘Ÿ^š‘˜ āˆ’ š‘Ž + š‘Žš‘Ÿ^š‘˜ (š‘Ÿ) āˆ’ š‘Žš‘Ÿ^š‘˜)/(š‘Ÿ āˆ’ 1) = (š‘Žš‘Ÿ^š‘˜āˆ’ š‘Žš‘Ÿ^š‘˜ āˆ’ š‘Ž + š‘Žš‘Ÿ^š‘˜ (š‘Ÿ))/(š‘Ÿ āˆ’ 1) = (0 āˆ’ š‘Ž + š‘Žš‘Ÿ^š‘˜ (š‘Ÿ))/(š‘Ÿ āˆ’ 1) = (āˆ’ š‘Ž + š‘Žš‘Ÿ^š‘˜ (š‘Ÿ))/(š‘Ÿ āˆ’ 1) = (āˆ’ š‘Ž + š‘Žš‘Ÿ^š‘˜ (š‘Ÿ^1 ))/(š‘Ÿ āˆ’ 1) = (āˆ’ š‘Ž + š‘Žš‘Ÿ^(š‘˜ + 1))/(š‘Ÿ āˆ’ 1) = (š‘Ž (āˆ’1 + š‘Ÿ^(š‘˜ + 1) ))/(š‘Ÿ āˆ’ 1) = š‘Ž(š‘Ÿ^(š‘˜ + 1) āˆ’ 1)/(š‘Ÿ āˆ’ 1) Thus, a + ar + ar2 + ……..+ ark – 1 + ark = š‘Ž(š‘Ÿ^(š‘˜ + 1) āˆ’ 1)/(š‘Ÿ āˆ’ 1) which is the same as P(k + 1) ∓ P(k + 1) is true whenever 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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