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(5328)   (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
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