Edward A. answered • 01/12/18

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High School Math Whiz grown up--I've even tutored my grandchildren

Proof by induction:

(1) prove the initial case

(2) show that if it is true through for n, it is also true for n+1

So (1) demonstrate for n =0

Evaluate proposed right hand side for n=0

(1) prove the initial case

(2) show that if it is true through for n, it is also true for n+1

So (1) demonstrate for n =0

Evaluate proposed right hand side for n=0

(q^1-1)/(q-1)

(q-1)/(q-1) =1

Left hand side

Q^0 =1

So it is true that the left hand side equals right hand side for n=0

(2) evaluate both sides for n+1 by adding the term q^(n+1)

Left hand side:

(Sum up to n+1) = (sum up to n) + q^(n+1)

Right hand side

q^(n+1)-1)/(q-1) + q^(n+1)

Combine over common denominator

((q^(n+1)-1) + q^(n+1)*(q-1)) /(q-1)

((q^(n+1)*(q-1+1) -1)) /(q-1)

((q^(n+1)*q -1))/(q-1)

((q^(n+2) -1))/(q-1)

This is the right hand side for the n+1 case

So the left hand side for n+1 equals the right hand side for n+1. So the induction step is true.

(q-1)/(q-1) =1

Left hand side

Q^0 =1

So it is true that the left hand side equals right hand side for n=0

(2) evaluate both sides for n+1 by adding the term q^(n+1)

Left hand side:

(Sum up to n+1) = (sum up to n) + q^(n+1)

Right hand side

q^(n+1)-1)/(q-1) + q^(n+1)

Combine over common denominator

((q^(n+1)-1) + q^(n+1)*(q-1)) /(q-1)

((q^(n+1)*(q-1+1) -1)) /(q-1)

((q^(n+1)*q -1))/(q-1)

((q^(n+2) -1))/(q-1)

This is the right hand side for the n+1 case

So the left hand side for n+1 equals the right hand side for n+1. So the induction step is true.