Thread: inductive math (I'm missing something)

fn is the nth Fibonacci number

Prove that f1(^2)+f2(^2)+...+fn(^2)=fn fn+1 whenever n is a positive integer.

Here's what I have so far:

Let P(n) be f1(^2)+f2(^2)+...fn(^2)=fn fn+1

Basis step: P(1) is true since f1(2)=1=f2(^2)

Inductive Step: Assume P(n)) is true
Then f1(^2)+f2(^2)+...fn(^2)+fn+1(^2)=fn fn+1 + fn+1(^2)


Have I done this right so far, and if so, where do I go from here?

Originally Posted by sjenkins

fn is the nth Fibonacci number

Prove that f1(^2)+f2(^2)+...+fn(^2)=fn fn+1 whenever n is a positive integer.

Here's what I have so far:

Let P(n) be f1(^2)+f2(^2)+...fn(^2)=fn fn+1

Basis step: P(1) is true since f1(2)=1=f2(^2)

this is wrong. you haven't verified the statement holds for the base case

Inductive Step: Assume P(n)) is true
Then f1(^2)+f2(^2)+...fn(^2)+fn+1(^2)=fn fn+1 + fn+1(^2)


Have I done this right so far, and if so, where do I go from here?

Hint:

(and in general, you get the next Fibonacci number by adding the previous two, this will come in handy)