Question

Can someone help me please to solve this?
\[\sum_{k=1}^{n} f _{k} = f _{n+2} = 1\]
In this problem \[\f _{k}\] denotes the nth Fibonacci number

Answer 1

hi can u help me please?

Answer 2

hi mathmate

Answer 3

were u able to finish that problem?

Answer 4

Not sure if you have a typo or not.
Have you done proof by induction?

Answer 5

I don't know how to do that

Answer 6

I was working on that problem, and tried to absorb the right hand-side into the constants, but I don't get the right answer, so still working on it.

Answer 7

i made a typo with the \[f _{k}\]...there shouldn't be a backslash before it

Answer 8

oh ok

Answer 9

the instruction is saying to prove \[\sum_{k=1}^{n} f _{k} = f _{n+2} = 1\]

Answer 10

I see two equal signs, is that correct?

Answer 11

yes

Answer 12

oh wait

Answer 13

i think the teacher made a typo

Answer 14

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

Answer 15

To prove:

fi=1,1,2,3,5,8,13,21,...
This can be proved by induction.

Base case: n=1
\[\sum_{1}^{1} f _{i} = 1\]
\[f _{3}-1 = 2-1 =1\]
Proposition works for n=1.

Answer 16

Induction step:
Assuming proposition is true for n=k,
then
\[\sum_{1}^{k} = f _{k+2} - 1\]
Then for k+1, add fk+1 to each side to get
\[\sum_{1}^{k}i + f _{k+1} = f _{k+2} + f _{k+1} -1\]

\[\sum_{1}^{k+1}i = f _{k+3} -1\] QED

Answer 17

QED?

Answer 18

"quod erat demonstrandum"
latin for what we would like to demonstrate.

Answer 19

oh ok

Answer 20

thank you

Answer 21

You're welcome.
Will get back to the recurrence relation.

Answer 22

ok no problem :)

Answer 23

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

Answer 24

It's the same prniciple as before:

Base case: n=1
\[\sum_{i=1}^{}f _{i} = f _{1} = 1\]
\[f _{2*1} = f _{2} =1\]
therefore proposition valid for n=1.

Assume proposition valid for n=k, then
\[\sum_{i=1}^{k}f _{2i-1} = f _{2k}\]
add the next term to the left-hand side:
\[\sum_{i=1}^{k}f _{2i-1} + f _{2k+1} = \sum_{i=1}^{k+1}f _{2i-1} \]
add the same term to the right hand side:
\[ f _{2k} + f _{2k+1} = f _{2k+2} \]
by definition of the fibonacci sequnce.
Therefore the proposition is valid also for n=k+1, therefore the proposition is valid for all integers.
QED