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1+r+r^2+r^3+...+r^n-1=1-r^n/1-r for all positive integer n. by mathematical induction. 

 Jan 7, 2017
 #1
avatar+9665 
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LHS stands for Left Hand Side and RHS stands for RHS. Use whatever notation your teacher tell you to use.

 

When n = 1,

LHS = 1

RHS = 1

LHS = RHS.

 

Assume that when n = k, LHS = RHS.

When n = k + 1,

\(LHS\\ = 1 + r + r^2 +... + r^k \\ =\dfrac{1-r^k}{1-r} + r^k\\ =\dfrac{1-r^k}{1-r} + \dfrac{r^k-r^{k+1}}{1-r}\\ =\dfrac{1-r^{k+1}}{1-r}\\ =RHS\)

 

\(\therefore\)1+r+r^2+r^3+...+r^n-1=1-r^n/1-r is true for all positive integer n.

 

~The smartest cookie in the world

 Jan 7, 2017
edited by MaxWong  Jan 7, 2017
 #2
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Proof: (Induction) Basis: Show true for n = 0. LHS = 1. RHS = [r (0+1) - 1]/(r - 1) = (r-1)/(r-1) = 1. Therefore LHS = RHS. Induction: Assume 1+r+r 2+...+r k = r (k+1)-1/r-1. Show 1+r+r 2+...+r k+r (k+1) = [r (k+2) - 1]/(r - 1). Now, 1+r+r 2+...+r k+r (k+1) = r (k+1)-1/r-1 + r (k+1) = [r (k+1) - 1 + (r - 1)r (k+1) ]/(r - 1) = [r (k+1) - 1 + r×r (k+1) - r (k+1) ]/(r - 1) = [r (k+2) - 1]/(r - 1). QED

 Jan 7, 2017
 #3
avatar+9665 
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We cannot show true for this for n = 0.....

If we substitute n = 0 in the LHS the last term of LHS will become \(\dfrac{1}{r}\).

 

So we show that this is true for n = 1......

MaxWong  Jan 7, 2017

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