# SOLUTION: Prove a^n -1=(a-1)(a^(n-1) + a^(n-2) +.......+ a+1) by using Principle of Mathematical Induction.

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 Question 633227: Prove a^n -1=(a-1)(a^(n-1) + a^(n-2) +.......+ a+1) by using Principle of Mathematical Induction.Answer by KMST(1874)   (Show Source): You can put this solution on YOUR website!a^n -1=(a-1)(a^(n-1) + a^(n-2) +.......+ a+1) We need to prove that it is true for n=1 or n=2,and we need to prove that if it is true for n=k, it must be true for n=k+1. is easy to prove for n=1 or for n=2 For n=1, a^(n-1)=a^(1-1)=a^0=1, and the long sum in parenthesis, going from a^0=1 down to a^0=1, is just 1. Substituting n=1 in a^n -1=(a-1)(a^(n-1) + a^(n-2) +.......+ a+1), we get a^1 - 1=(a-1)(1), which is trivial. For n=2, substituting we get a^2 - 1=(a-1)(a+1) which we know is true. (It's one of those special products we learn for factoring). If we assume that a^n - 1=(a-1)(a^(n-1) + a^(n-2) +.......+ a+1) is true for any positive integer , a^k - 1=(a-1)(a^(k-1) + a^(k-2) +.......+ a+1) Adding 1 to both sides, we get a^k =(a-1)(a^(k-1) + a^(k-2) +.......+ a+1)+1 Multiplying both sides times a, we get a*a^k =((a-1)(a^(k-1) + a^(k-2) +.......+ a+1)+1)a Distributing a^(k+1) =(a-1)(a^(k-1) + a^(k-2) +.......+ a+1)*a+1*a Applying the associative property a^(k+1) =(a-1)((a^(k-1) + a^(k-2) +.......+ a+1)*a}+1*a Distributing a^(k+1) =(a-1)(a^(k-1+1) + a^(k-2+1) +.......+ a*a+1*a)+a a^(k+1) =(a-1)(a^k + a^(k-1) +.......+ a^2+a)+a a^(k+1)-1 =(a-1)(a^k + a^(k-1) +.......+ a^2+a)+a-1 a^(k+1)-1 =(a-1)(a^k + a^(k-1) +.......+ a^2+a)+(a-1) a^(k+1)-1 =(a-1)((a^k + a^(k-1) +.......+ a^2+a)+1) a^(k+1)-1 =(a-1)(a^k + a^(k-1) +.......+ a^2+a+1) which is a^n -1=(a-1)(a^(n-1) + a^(n-2) +.......+ a+1) for n=k+1