# fi is fibonacci number in order to i Prove... 1^2+1^2+2^2+3^2+5^2+...+fn^2 = fn+1 x fn n+1, n are index of the numbers.Could anyone prove this? fi is fibonacci number in order to i Prove......

fi is fibonacci number in order to i Prove... 1^2+1^2+2^2+3^2+5^2+...+fn^2 = fn+1 x fn n+1, n are index of the numbers.

Could anyone prove this?

### n+1, n are index of the numbers.

Could anyone prove this?

neela | Student

To prove that 1^2+1^2+2^2+3^2+5^2+...+fn^2 = fn+1 x fn, where n = 1, 2,3,....

We notice that the terms the fibinocci are as follows:

f1 = f2 = 1,

f3 = f2+f1 = 1+1 =2,

f4 = f3+f2 = 2+1 =3,

f5= f4+f3= 5,

Similarly fn = fn-1+fn-2.

Therefore fn+1 = fn+fn-1

So fn+1*fn = (fn+fn-1)fn.

=> fn*fn+1 = fn^2-fn*fn-1

So we got the recurrence relation fn*fn+1 = fn^2+fn*fn-1.

fn+1*fn = fn^2+fn*fn-1.

Applying the recurrence relation for the second tern fn*fn-1 we get:

fn+1*fn = fn^2+fn-1^2+fn-1*fn-2.

Similarly applying the recurrence relation successively we get:

fn+1*fn = fn^2+fn-1^2+fn-2^2+....f3^2+f2^2+f2*f1.

fn+1*fn = fn^2+fn-1^2+fn-2^2+....f3^2+f2^2+f1^2, as f2 = f1.

Therefore fn+1*fn = fn^2+fn-1^2 +....21^2+13^2+8^2+5^2+3^2+2^2+1^2+1^2.

Therefore 1^2+1^2+3^2+5^2+8^2+13^2+21^2+....+fn^2 = fn*fn+1.

yuniahoy | Student

ohhh. awesome... ^^

yuniahoy | Student

could you help me again?..

I have one difficult question from my teacher..

5^x - 6squareX + 5 = 0

Determine X value..

Thanks..^^

yuniahoy | Student

ohhh. awesome... ^^