# How to postulate a formula for the following Mathematica function

I am using Mathematica to explore the properties of the Fibonacci Sequence. Below is the functions I have defined

H[n_] := \[Phi]^n/Sqrt

F[n_] := Round[H[n]]

S[m_] := Sum[F[n], {n, 1, m}]


Through experimentation I have realised that S[m]=SF[m-1]+F[m]. How do I obtain a conjectured formula for this and prove it by induction?

Please let me know if you need me to clarify anything further.

• Are you aware that Fibonacci[] is built-in? Try this: FullSimplify[Sum[Fibonacci[k], {k, 1, n}] == Fibonacci[n + 1] + Fibonacci[n] - 1, Element[n, Integers]] – J. M. is in limbo Oct 11 '18 at 15:42
• What is SF[m]? – Daniel Lichtblau Oct 12 '18 at 10:12

H[n_] = GoldenRatio^n/Sqrt;

S[m_] := Total@Round[H /@ Range[m] // FunctionExpand // Simplify]


Generate a sequence of S[m]

seq = S /@ Range

(* {1, 2, 4, 7, 12, 20, 33, 54, 88, 143} *)


Use FindSequenceFunction to find a function that generates the sequence

S2[m_] = FindSequenceFunction[seq, m]

(* 1/2 (-2 + 3 Fibonacci[m] + LucasL[m]) *)


Verifying that S and S2 are equivalent beyond the range of the original sequence

And @@ (Table[S[m] == S2[m], {m, 100}] // Simplify)

(* True *)


While S[m] is discrete, S2[m] is continuous.

Show[
Plot[S2[m], {m, 0, 5.25}],
DiscretePlot[S[m], {m, 0, 5}, PlotStyle -> Red]] S2[m] == S2[m - 1] + Fibonacci[m] // FunctionExpand // Simplify

• Fibonacci[n + 2] - 1 would be a much simpler expression for the sum; the Identity for S2 then reduces to the defining recursion for the Fibonacci numbers. – J. M. is in limbo Oct 12 '18 at 1:32