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
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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