Verified answer

Sui, there is a slight error in your problem.

Let us say Sn = Σ (fk)(fk+1) (k goes from 1 to 2n)

We have f1=0, f2=1, f3=1, f4=2, f5=3, f6=5, f7=8.

for n = 3 S3 = f1f2 + f2f3 + f3f4 + f4f5 +f5f6 + f6f7

S3 = 0+1+2+6+15+40=64 = f7²

Your problem should then read Sn = (f2n+1)².

We showed it is true for n=3. (for n=1, S1=f1f2+f2f3=1=f3² and for n=2 S2 = f1f2+f2f3+f3f4+f4f5 = 0+1+2+6 = 9 = f5²)

Let us prove it is true for n+1:

S(n+1) = Σ (fk)(fk+1) (for k=1 to 2n+2)

S(n+1) = Sn + (f2n+1)(f2n+2) + (f2n+2)(f2n+3)

S(n+1) = (f2n+1)² + (f2n+1)(f2n+2) + (f2n+2)((f2n+1)+(f2n+2))

S(n+1) = (f2n+1)² + 2(f2n+1)(f2n+2) + (f2n+2)²

S(n+1) = ((f2n+1) + (f2n+2))²

S(n+1) = (f2n+3)²

It is therefore true for every n>0

Edit: if we say f1=1, f2=1, f3=2, f4=3, f5=5, f6=8, f7=13,

then S3 = f1f2 +f2f3 +f3f4 +f4f5 + f5f6 + f6f7 = 168, not a perfect square.

We'd have to say S3 = f1f2 +f2f3 +f3f4 +f4f5 + f5f6 = 64 =8² = f6² but then Sn = Σ (fk)(fk-1) (k goes from 1 to 2n)

The proof is basically the same.

(Sn+1) = Sn + (f2n)(f2n+1) + (f2n+1)(f2n+2)

(Sn+1) = f2n² + (f2n)(f2n+1) + (f2n+1)((f2n+1) + f2n)

(Sn+1) = f2n² + 2(f2n)(f2n+1) + (f2n+1)²

(Sn+1) = (f2n + (f2n+1))² = (f2n+2)²

It all boils down to where you start the fibonacci sequence.

Iif you decide f1=1 then your problem should read Σ (fk)(fk-1) (k goes from 1 to 2n) = f2n²

If you decide f1=0 your problem should read Σ (fk)(fk+1) (k goes from 1 to 2n) = (f2n+1)²

The proof is identical either way.