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

  1. Chapter 4 Class 11 Mathematical Induction
  2. Serial order wise
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Ex 4.1, 12: Prove the following by using the principle of mathematical induction for all n โˆˆ N: a + ar + ar2 + โ€ฆโ€ฆ..+ arn โ€“ 1 = (๐‘Ž(๐‘Ÿ^๐‘› โˆ’ 1))/(๐‘Ÿ โˆ’ 1) Let P (n) : a + ar + ar2 + โ€ฆโ€ฆ..+ arn โ€“ 1 = ๐‘Ž(๐‘Ÿ^๐‘› โˆ’ 1)/(๐‘Ÿ โˆ’ 1) For n = 1, L.H.S = a R.H.S = (๐‘Ž(๐‘Ÿ1 โˆ’ 1))/(๐‘Ÿ โˆ’ 1) = (๐‘Ž(๐‘Ÿ โˆ’ 1))/(๐‘Ÿ โˆ’ 1) = a L.H.S. = R.H.S โˆด P(n) is true for n = 1 Assume that P(k) is true a + ar + ar2 + โ€ฆโ€ฆ..+ ark โ€“ 1 = ๐‘Ž(๐‘Ÿ^๐‘˜ โˆ’ 1)/(๐‘Ÿ โˆ’ 1) We will prove that P(k + 1) is true. a + ar + ar2 + โ€ฆโ€ฆ..+ ar(k + 1) โ€“ 1 = ๐‘Ž(๐‘Ÿ^(๐‘˜ + 1) โˆ’ 1)/(๐‘Ÿ โˆ’ 1) a + ar + ar2 + โ€ฆโ€ฆ..+ ark โ€“ 1 + ark = ๐‘Ž(๐‘Ÿ^(๐‘˜ + 1) โˆ’ 1)/(๐‘Ÿ โˆ’ 1) We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) a + ar + ar2 + โ€ฆโ€ฆ..+ ark โ€“ 1 = ๐‘Ž(๐‘Ÿ^๐‘˜ โˆ’ 1)/(๐‘Ÿ โˆ’ 1) Adding ark both sides a + ar + ar2 + โ€ฆโ€ฆ.. +ark โ€“ 1 + ark = ๐‘Ž(๐‘Ÿ^๐‘˜ โˆ’ 1)/(๐‘Ÿ โˆ’ 1) + ark = (๐‘Ž(๐‘Ÿ^๐‘˜ โˆ’ 1) + (๐‘Ÿ โˆ’ 1)๐‘Ž๐‘Ÿ^๐‘˜)/(๐‘Ÿ โˆ’ 1) = (๐‘Ž๐‘Ÿ^๐‘˜ โˆ’ ๐‘Ž + ๐‘Ž๐‘Ÿ^๐‘˜ (๐‘Ÿ) โˆ’ ๐‘Ž๐‘Ÿ^๐‘˜)/(๐‘Ÿ โˆ’ 1) = (๐‘Ž๐‘Ÿ^๐‘˜โˆ’ ๐‘Ž๐‘Ÿ^๐‘˜ โˆ’ ๐‘Ž + ๐‘Ž๐‘Ÿ^๐‘˜ (๐‘Ÿ))/(๐‘Ÿ โˆ’ 1) = (0 โˆ’ ๐‘Ž + ๐‘Ž๐‘Ÿ^๐‘˜ (๐‘Ÿ))/(๐‘Ÿ โˆ’ 1) = (โˆ’ ๐‘Ž + ๐‘Ž๐‘Ÿ^๐‘˜ (๐‘Ÿ))/(๐‘Ÿ โˆ’ 1) = (โˆ’ ๐‘Ž + ๐‘Ž๐‘Ÿ^๐‘˜ (๐‘Ÿ^1 ))/(๐‘Ÿ โˆ’ 1) = (โˆ’ ๐‘Ž + ๐‘Ž๐‘Ÿ^(๐‘˜ + 1))/(๐‘Ÿ โˆ’ 1) = (๐‘Ž (โˆ’1 + ๐‘Ÿ^(๐‘˜ + 1) ))/(๐‘Ÿ โˆ’ 1) = ๐‘Ž(๐‘Ÿ^(๐‘˜ + 1) โˆ’ 1)/(๐‘Ÿ โˆ’ 1) Thus, a + ar + ar2 + โ€ฆโ€ฆ..+ ark โ€“ 1 + ark = ๐‘Ž(๐‘Ÿ^(๐‘˜ + 1) โˆ’ 1)/(๐‘Ÿ โˆ’ 1) which is the same as P(k + 1) โˆด P(k + 1) is true whenever P(k) is true. โˆด By the principle of mathematical induction, P(n) is true for n, where n is a natural number

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