SOLUTION: Prove using simple induction that, for each integer n >= 1 3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2

Question 736005: Prove using simple induction that, for each integer n >= 1
3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Prove using simple induction that, for each integer n >= 1
3 + 3^2+ 3^3+...+ 3^n = (3^(n+1)-3)/2
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Show it is true for n = 1
3 = (3^(1+1)-3)/2 = (3^2 - 3)/2 = 6/2 = 3
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Assume it is true for n = k:
3 + 3^2 +...+3^k = (3^(k+1) -3)/2
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Show it must be true for n = k+1
[3 + 3^2+...+3^k] + 3^(k+1)
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= (3^(k+1) -3)/2 + 3^(k+1)
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= [3^(k+1) - 3) + 2*3^(k+1)]/2
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= [3*3^(k+1) - 3]/2
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= [3^((k+1)+1) - 3]/2
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Cheers,
Stan H.
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