Question

Need help proving

\[\sum_{k=1}^{n}F ^{2}_{k}-F _{n}F _{n+1}\]

by induction.

Answer 1

what is it supposed to equal?

Answer 2

that - is supposed to be =
sorry

Answer 3

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 4

ok we can do this i think. have you done if for the case of n = 1?

Answer 5

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 6

Yes. F1 is 1 and Fn+1 is also 1 so it works.

Answer 7

ok so you want to assume that it is true for say n and show that it is true for n + 1
is that how you wish to proceed?

Answer 8

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 9

right, by induction

Answer 10

or would you like to assume it is true for n - 1 and show that it is true for n?

Answer 11

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 12

n+1 is how we've been doing it in class, so that's how I should probably do it.

Answer 13

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 14

I just saw Joe's attachment. It looks like what I need.

Answer 15

lets do it the first way. assume it is true for n, i.e. assume that we know
\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\] and now consider
\[F_{n+1}F_{n+2}=F_{n+1}(F_{n+1}+F_n)=F_{n+1}^2+F_{n+1}F_n\]

Answer 16

by assumption the second term is
\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\] so the whole thing is
\[\sum_{k=1}^{n+1}F_k^2\]

Answer 17

oops missed joe's attachment!

Answer 18

missed joe too! how you been?

Answer 19

im not doing too good =/ really sick, and my school work is above my head.

Answer 20

sorry to hear that. have some chicken soup. get some rest. and stop wasting time when you should be in bed.

Answer 21

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 22

Hi again Joe, thanks to you and Satellite both. I have a question though, your base step begins with 0. I thought the Fib sequences started with 2 1's?

Answer 23

you can start as 1, 1, 2, ..

Answer 24

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 25

ok... that's how I understood it...
It looks like it works.

Answer 26

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 27

Thanks again to both of you for your help. I greatly appreciate it.

Answer 28

i start with 0, 1.

It makes the closed formula for the sequence come out nicer.

Answer 29

\[\sum_{k=1}^{n}F ^{2}_{k}=F _{n}F _{n+1}\]

Answer 30

Yeah, I think he told us that it can be done that way... the book we use starts with the ones though and so that's how he presented it in class and in examples. Your proof makes sense to me though. Thanks :)

Answer 31

yw (for joe too)