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Last updated at May 29, 2018 by Teachoo
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Ex 4.1, 12: Prove the following by using the principle of mathematical induction for all n โ N: a + ar + ar2 + โฆโฆ..+ arn โ 1 = (๐(๐^๐ โ 1))/(๐ โ 1) Let P (n) : a + ar + ar2 + โฆโฆ..+ arn โ 1 = ๐(๐^๐ โ 1)/(๐ โ 1) For n = 1, L.H.S = a R.H.S = (๐(๐1 โ 1))/(๐ โ 1) = (๐(๐ โ 1))/(๐ โ 1) = a L.H.S. = R.H.S โด P(n) is true for n = 1 Assume that P(k) is true a + ar + ar2 + โฆโฆ..+ ark โ 1 = ๐(๐^๐ โ 1)/(๐ โ 1) We will prove that P(k + 1) is true. a + ar + ar2 + โฆโฆ..+ ar(k + 1) โ 1 = ๐(๐^(๐ + 1) โ 1)/(๐ โ 1) a + ar + ar2 + โฆโฆ..+ ark โ 1 + ark = ๐(๐^(๐ + 1) โ 1)/(๐ โ 1) We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) a + ar + ar2 + โฆโฆ..+ ark โ 1 = ๐(๐^๐ โ 1)/(๐ โ 1) Adding ark both sides a + ar + ar2 + โฆโฆ.. +ark โ 1 + ark = ๐(๐^๐ โ 1)/(๐ โ 1) + ark = (๐(๐^๐ โ 1) + (๐ โ 1)๐๐^๐)/(๐ โ 1) = (๐๐^๐ โ ๐ + ๐๐^๐ (๐) โ ๐๐^๐)/(๐ โ 1) = (๐๐^๐โ ๐๐^๐ โ ๐ + ๐๐^๐ (๐))/(๐ โ 1) = (0 โ ๐ + ๐๐^๐ (๐))/(๐ โ 1) = (โ ๐ + ๐๐^๐ (๐))/(๐ โ 1) = (โ ๐ + ๐๐^๐ (๐^1 ))/(๐ โ 1) = (โ ๐ + ๐๐^(๐ + 1))/(๐ โ 1) = (๐ (โ1 + ๐^(๐ + 1) ))/(๐ โ 1) = ๐(๐^(๐ + 1) โ 1)/(๐ โ 1) Thus, a + ar + ar2 + โฆโฆ..+ ark โ 1 + ark = ๐(๐^(๐ + 1) โ 1)/(๐ โ 1) which is the same as P(k + 1) โด P(k + 1) is true whenever P(k) is true. โด By the principle of mathematical induction, P(n) is true for n, where n is a natural number
Ex 4.1
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