SOLUTION: Prove by induction: 1+a+a^2+...+a^(n-1)=(a^n-1)/(a-1)

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Question 962542: Prove by induction:
1+a+a^2+...+a^(n-1)=(a^n-1)/(a-1)
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1%2Ba%2Ba%5E2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Ba%5E%28n-1%29=%28a%5En-1%29%2F%28a-1%29

Prove true for n=1: (There is only one term)

1=a%5E%281-1%29
1=a%5E0
1=1

So it is true for n=1

Assume it is true for all n ≤ k

1%2Ba%2Ba%5E2%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2Ba%5E%28k-1%29=%28a%5Ek-1%29%2F%28a-1%29

Add a%5Ek to both sides



Simplifying the right side:

%28a%5Ek-1%29%2F%28a-1%29%2Ba%5Ek

%28a%5Ek-1%29%2F%28a-1%29%2Ba%5Ek%2F1

%28a%5Ek-1%29%2F%28a-1%29%2Ba%5Ek%2A%28a-1%29%2F%28a-1%29

%28%28a%5Ek-1%29%2Ba%5Ek%2A%28a-1%29%29%2F%28a-1%29

%28a%5Ek-1%2Ba%5E%28k%2B1%29-a%5Ek%29%2F%28a-1%29 

%28cross%28a%5Ek%29-1%2Ba%5E%28k%2B1%29-cross%28a%5Ek%29%29%2F%28a-1%29 

%28-1%2Ba%5E%28k%2B1%29%29%2F%28a-1%29

%28a%5E%28k%2B1%29-1%29%2F%28a-1%29

And that is exactly what you would get if you substitute
n=k+1 in %28a%5En-1%29%2F%28a-1%29.

So the formula is proved.

Edwin