Prove that for any four consecutive Fibonacci numbers a, b, c, d, the square root of a^2 + 4bc...

is always a perfect square whose square root is equal to d.To prove that for any four consecutive Fibonacci numbers a, b, c, d, the square root of a^2 + 4bc is equal to d.

For any four consecutive Fibonacci numbers a, b, c, d,

a = c - b ............(equation i)

By substituting for a in a^2 + 4bc we have:

a^2 + 4bc

= (c-b)^2 + 4bc

= c^2 -2bc + b^2 + 4bc

= c^2+2bc + b^2

= (c+b)^2

Therefore, a^2 + 4bc = (c+b)^2.

Taking the square root of both sides we have:

The square root of a^2 + 4bc is equal to c+b or b+c.

But b+c = d for any three consecutive Fibonacci numbers b, c, d.

Therefore, square root of a^2 + 4bc is equal to d.

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