Equal - Addition
Last updated at July 30, 2026 by Teachoo
Transcript
Question 12: Prove the following by using the principle of mathematical induction for all n ā N: a + ar + ar2 + ā¦ā¦..+ arn ā 1 = (š(š^š ā 1))/(š ā 1) Let P (n) : a + ar + ar2 + ā¦ā¦..+ arn ā 1 = š(š^š ā 1)/(š ā 1) For n = 1, L.H.S = a R.H.S = (š(š1 ā 1))/(š ā 1) = (š(š ā 1))/(š ā 1) = a L.H.S. = R.H.S ā“ P(n) is true for n = 1 Assume that P(k) is true a + ar + ar2 + ā¦ā¦..+ ark ā 1 = š(š^š ā 1)/(š ā 1) We will prove that P(k + 1) is true. a + ar + ar2 + ā¦ā¦..+ ar(k + 1) ā 1 = š(š^(š + 1) ā 1)/(š ā 1) a + ar + ar2 + ā¦ā¦..+ ark ā 1 + ark = š(š^(š + 1) ā 1)/(š ā 1) We have to prove P(k+1) from P(k) i.e. (2) from (1) From (1) a + ar + ar2 + ā¦ā¦..+ ark ā 1 = š(š^š ā 1)/(š ā 1) Adding ark both sides a + ar + ar2 + ā¦ā¦.. +ark ā 1 + ark = š(š^š ā 1)/(š ā 1) + ark = (š(š^š ā 1) + (š ā 1)šš^š)/(š ā 1) = (šš^š ā š + šš^š (š) ā šš^š)/(š ā 1) = (šš^šā šš^š ā š + šš^š (š))/(š ā 1) = (0 ā š + šš^š (š))/(š ā 1) = (ā š + šš^š (š))/(š ā 1) = (ā š + šš^š (š^1 ))/(š ā 1) = (ā š + šš^(š + 1))/(š ā 1) = (š (ā1 + š^(š + 1) ))/(š ā 1) = š(š^(š + 1) ā 1)/(š ā 1) Thus, a + ar + ar2 + ā¦ā¦..+ ark ā 1 + ark = š(š^(š + 1) ā 1)/(š ā 1) which is the same as P(k + 1) ā“ P(k + 1) is true whenever P(k) is true. ā“ By the principle of mathematical induction, P(n) is true for n, where n is a natural number