hcju/proofwikiqa huggingface dataset:
We have that: :$\displaystyle \sum_{k \mathop = 1}^n \dfrac 1 {2 k - 1} = \sum_{i \mathop = 0}^{M - 1} \paren {\dfrac 1 {i \times 2^{r + 1} + 1} + \dfrac 1 {i \times 2^{r + 1} + 3} + \cdots + \dfrac 1 {\paren {i + 1} \times 2^{r + 1} - 1} }$ where $k = 2^r M$ where $M$ is odd. Collect the $r$ terms in the parenthesis on the {{RHS}} of the $i$th term under a single common denominator $P_i$. Then the resulting numerators will each consist of $2^r$ terms, each of the form $\dfrac {P_i} {R_i}$, where $R_i$ is a distinct odd residue of $2^{r + 1}$. These $\dfrac {P_i} {R_i}$ must themselves also be the distinct odd residue of $2^{r + 1}$, in some order. The odd residues of $2^{r + 1}$ are: :$1, 3, 5, \dots, 2^{r + 1} - 1$ Their sum is: {{begin-eqn}} {{eqn | o =       | r = 1 + 3 + \cdots + \paren {2^{r + 1} - 1} }} {{eqn | r = \frac {\paren {1 + 2^{r + 1} - 1} 2^r} 2       | c = Sum of Arithmetic Sequence }} {{eqn | r = 2^{2 r} }} {{end-eqn}} Thus each of the $M$ numerators is of the form $2^{2 r} M_i$, where $M_i$ is odd. Thus the numerator of $S$ is in the form $2^{2 r} Q$ where $Q$ is the sum of an odd number of odd terms. Therefore $Q$ is itself odd. Hence, if $2^r$ is the largest power of $2$ which divides $k$, then $2^{2 r}$ is the largest power of $2$ dividing the numerator of $S$. {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^1 \dfrac {\map f {x_0} - \map f {x_0 y} } {1 - y} \rd y       | r = \int_0^1 \dfrac {\sum_{k \mathop \ge 0} a_k {x_0}^k - \sum_{k \mathop \ge 0} a_k {x_0}^k y^k} {1 - y} \rd y       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \dfrac {1 - y^k} {1 - y} \rd y       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \int_0^1 \paren {1 + y + y^2 + \cdots + y^{k - 1} } \rd y       | c = Sum of Geometric Sequence }} {{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \intlimits {y + \dfrac {y^2} 2 + \dfrac {y^3} 3 + \cdots + \dfrac {y^k} k} 0 1       | c = Primitive of Power }} {{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k \paren {\paren {1 + \dfrac {1^2} 2 + \dfrac {1^3} 3 + \cdots + \dfrac {1^k} k} - \paren {0 + \dfrac {0^2} 2 + \dfrac {0^3} 3 + \cdots + \dfrac {0^k} k} }       | c =  }} {{eqn | r = \sum_{k \mathop \ge 0} a_k {x_0}^k H_k       | c = {{Defof|Harmonic Numbers}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop = 1}^n \dfrac {H_k} {n + 1 - k}       | r = \sum_{k \mathop = 1}^n \dfrac {H_{n + 1 - k} } k       | c = Permutation of Indices of Summation }} {{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = 1}^k \dfrac 1 {j \paren {n + 1 - k} }       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \sum_{k \mathop = 1}^n \sum_{j \mathop = k}^n \dfrac 1 {\paren {n + 1 - j} k}       | c = {{Defof|Harmonic Number}} }} {{end-eqn}} {{ProofWanted|all at sea}}
For $n \in \N$: {{begin-eqn}} {{eqn | l = \map H n       | r = \gamma + \frac {\map {\Gamma'} {n + 1} } {\map \Gamma {n + 1} } }} {{eqn | r = \gamma - \gamma + \sum_{m \mathop = 1}^\infty \paren {\frac 1 m - \frac 1 {n + m} }       | c = Reciprocal times Derivative of Gamma Function }} {{eqn | r = \lim_{k \mathop \to \infty} \sum_{m \mathop = 1}^k \paren {\frac 1 m - \frac 1 {n + m} } }} {{eqn | r = \lim_{k \mathop \to \infty} \paren {\sum_{m \mathop = 1}^k \paren {\frac 1 m} - \sum_{m \mathop = 1}^k \paren {\frac 1 {n + m} } } }} {{eqn | r = \lim_{k \mathop \to \infty} \paren {\sum_{m \mathop = 1}^k \paren {\frac 1 m} - \sum_{m \mathop = n + 1}^{k + n} \paren {\frac 1 m } } }} {{eqn | r = \lim_{k \mathop \to \infty} \paren {\sum_{m \mathop = 1}^n \paren {\frac 1 m} - \sum_{m \mathop = k + 1}^{k + n} \paren {\frac 1 m } } }} {{eqn | r = \sum_{m \mathop = 1}^n \paren {\frac 1 m} - \lim_{k \mathop \to \infty} \sum_{m \mathop = k + 1}^{k + n} \paren {\frac 1 m }       | c = this limit will vanish since it is positive but less than $\dfrac n {k + 1}$ }} {{eqn | r = H_n - 0       | c = {{Defof|Harmonic Number}} }} {{eqn | r = H_n }} {{end-eqn}} {{qed}}
There are no elements in the domain of $\nu$. Thus $\nu$ is an injection vacuously. {{qed}} Category:Injections Category:Empty Set 082mtnspem1bu64j5z32dcoiqwfnjxn
From Empty Mapping is Injective, $\nu$ is injective. As the codomain of $\nu$ is empty, $\nu$ is vacuously surjective. {{qed}} Category:Bijections Category:Empty Set lg8zos0rnftsvotoyqp8z32dt2hv6d6
From Fibonacci Number greater than Golden Section to Power less Two: :$F_n \ge \phi^{n - 2}$ From Fibonacci Number less than Golden Section to Power less One: :$F_n \le \phi^{n - 1}$ {{qed}}
Let the form of $\map G z$ be assumed as: {{begin-eqn}} {{eqn | l = \map G z       | r = \sum_{k \mathop \ge 0} F_k z^k       | c =  }} {{eqn | r = F_0 + F_1 z + F_2 z^2 + F_3 z^3 + F_4 z^4 + \cdots       | c =  }} {{eqn | r = 0 + z + z^2 + 2 z^3 + 3 z^4 + \cdots       | c =  }} {{end-eqn}} where $F_n$ denotes the $n$th Fibonacci number. Then: {{begin-eqn}} {{eqn | l = z \map G z       | r = F_0 z + F_1 z^2 + F_2 z^3 + F_3 z^4 + F_4 z^5 + \cdots       | c =  }} {{eqn | l = z^2 \map G z       | r = F_0 z^2 + F_1 z^3 + F_2 z^4 + F_3 z^5 + F_4 z^6 + \cdots       | c =  }} {{end-eqn}} and so: {{begin-eqn}} {{eqn | l = \paren {1 - z - z^2} \map G z       | r = F_0 + \paren {F_1 - F_0} z + \paren {F_2 - F_1 - F_0} z^2 + \paren {F_3 - F_2 - F_1} z^3 + \cdots       | c =  }} {{eqn | r = F_0 + \paren {F_1 - F_0} z       | c = {{Defof|Fibonacci Number}}: $F_n = F_{n - 1} + F_{n - 2}$ }} {{eqn | r = z       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{end-eqn}} Hence the result: :$\map G z = \dfrac z {1 - z - z^2}$ {{qed}}
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is: :$G \left({z}\right) = \dfrac z {1 - z - z^2}$ Then: {{begin-eqn}} {{eqn | l = G \left({z}\right)       | r = \dfrac z {1 - z - z^2}       | c =  }} {{eqn | r = \dfrac 1 {\sqrt 5} \left({\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }\right)       | c = Partial Fraction Expansion }} {{end-eqn}} where: :$\phi = \dfrac {1 + \sqrt 5} 2$ :$\hat \phi = \dfrac {1 - \sqrt 5} 2$ Hence: {{begin-eqn}} {{eqn | l = \left({G \left({z}\right)}\right)^2       | r = \left({\dfrac 1 {\sqrt 5} \left({\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }\right)}\right)^2       | c =  }} {{eqn | r = \dfrac 1 5 \left({\left({\dfrac 1 {1 - \phi z} }\right)^2 + \left({\dfrac 1 {1 - \hat \phi z} }\right)^2 - 2 \dfrac 1 {1 - \phi z} \dfrac 1 {1 - \hat \phi z} }\right)       | c =  }} {{eqn | r = \dfrac 1 5 \left({\dfrac 1 {\left({1 - \phi z}\right)^2} + \dfrac 1 {\left({1 - \hat \phi z}\right)^2} - 2 \dfrac {1 - \hat \phi z + 1 - \phi z} {\left({1 - \phi z}\right) \left({1 - \hat \phi z}\right)} }\right)       | c =  }} {{eqn | r = \dfrac 1 5 \left({\dfrac 1 {\left({1 - \phi z}\right)^2} + \dfrac 1 {\left({1 - \hat \phi z}\right)^2} - \dfrac 2 {\left({1 - \phi z}\right) \left({1 - \left({1 - \phi}\right) z}\right)} }\right)       | c = Definition of $\hat \phi$ }} {{eqn | r = \dfrac 1 5 \left({\dfrac 1 {\left({1 - \phi z}\right)^2} + \dfrac 1 {\left({1 - \hat \phi z}\right)^2} - \dfrac 2 {1 - z - z^2} }\right)       | c = after algebra }} {{end-eqn}} Thus the coefficient of $z^n$ in $\left({G \left({z}\right)}\right)^2$ is $\displaystyle \sum_{k \mathop = 0}^n F_k F_{n - k}$. {{explain|More detail needed}} Hence: {{begin-eqn}} {{eqn | l = \sum_{k \mathop = 0}^n F_k F_{n - k}       | r = \dfrac 1 5 \left({\left({n + 1}\right) \left({\phi^n + \hat \phi^n}\right) - 2 F_{n + 1} }\right)       | c =  }} {{eqn | r = \dfrac 1 5 \left({\left({n + 1}\right) \left({F_n + 2 F_{n - 1} }\right) - 2 F_{n + 1} }\right)       | c =  }} {{eqn | r = \dfrac {\left({n - 1}\right) F_n + 2n F_{n - 1} } 5       | c =  }} {{end-eqn}} {{qed}}
By definition of the Fibonacci numbers: {{begin-eqn}} {{eqn | l = F_0       | r = 0 }} {{eqn | l = F_1       | r = 1 }} {{end-eqn}} Then it is observed that $F_5 = 5$. After that, for $n > 5$, we have that $F_n > n$. {{qed}}
By definition of the Fibonacci numbers: {{begin-eqn}} {{eqn | l = F_0       | r = 0 }} {{eqn | l = F_1       | r = 1 }} {{end-eqn}} Then it is observed that $F_{12} = 144$. After that, for $n > 12$, we have that $F_n > n^2$. {{qed}}
Let $n \in \Z_{> 0}$. Then: {{begin-eqn}} {{eqn | l = F_{-\left({n + 1}\right)} F_{-\left({n - 1}\right)} - {F_{-n} }^2       | r = \left({-1}\right)^{n + 2} F_{n + 1} \left({-1}\right)^n F_{n - 1} - \left({\left({-1}\right)^{n + 1} F_n}\right)^2       | c = Fibonacci Number with Negative Index }} {{eqn | r = \left({-1}\right)^{2 n + 2} F_{n + 1} F_{n - 1} - \left({-1}\right)^{2 n + 2} F_n^2       | c =  }} {{eqn | r = F_{n + 1} F_{n - 1} - F_n^2       | c = $-1$ to an even power is $1$ }} {{end-eqn}} {{qed}}
Let $n \in \Z_{> 0}$. Then: {{begin-eqn}} {{eqn | l = \dfrac {\phi^{-n} - \hat \phi^{-n} } {\sqrt 5}       | r = \dfrac 1 {\sqrt 5} \paren {\phi^{-n} - \paren {-\dfrac 1 \phi}^{-n} }       | c = Definition of $\hat \phi$ }} {{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {\phi^n} - \paren {-1}^n \phi^n}       | c = Exponent Combination Laws: Negative Power }} {{eqn | r = \dfrac 1 {\sqrt 5} \paren {\paren {-1}^n \paren {-\dfrac 1 \phi}^n - \paren {-1}^n \phi^n}       | c =  }} {{eqn | r = \paren {-1}^n \dfrac 1 {\sqrt 5} \paren {\hat \phi^n - \phi^n}       | c = Definition of $\hat \phi$ }} {{eqn | r = \paren {-1}^{n + 1} \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}       | c =  }} {{eqn | r = \paren {-1}^{n + 1} F_n       | c = Euler-Binet Formula for positive $n$ }} {{eqn | r = F_{-n}       | c = Fibonacci Number with Negative Index }} {{end-eqn}} {{qed}}
=== Positive Index === First the result is proved for positive integers. {{:Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Positive Index}}{{qed|lemma}} === Negative Index === Then the result is extended to negative integers. {{:Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less/Negative Index}}{{qed|lemma}} Hence the result is seen to show for both positive and negative integers. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\phi^n = F_n \phi + F_{n - 1}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_0 \times \phi + F_{-1}       | r = F_0 \times \phi + \left({-1}\right)^0 F_1       | c = Fibonacci Number with Negative Index }} {{eqn | r = F_0 \times \phi + 1       | c = {{Defof|Fibonacci Number}} $F_1 = 1$ }} {{eqn | r = 0 \times \phi + 1       | c = {{Defof|Fibonacci Number}} $F_0 = 0$ }} {{eqn | r = 1       | c =  }} {{eqn | r = \phi^0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. ==== Basis for the Induction ==== $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_1 \times \phi + F_0       | r = F_1 \times \phi       | c = {{Defof|Fibonacci Number}} $F_0 = 0$ }} {{eqn | r = 1 \times \phi       | c = {{Defof|Fibonacci Number}} $F_1 = 1$ }} {{eqn | r = \phi^1       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. ==== Induction Hypothesis ==== Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\phi^k = F_k \phi + F_{k - 1}$ from which it is to be shown that: :$\phi^{k + 1} = F_{k + 1} \phi + F_k$ ==== Induction Step ==== This is the induction step: {{begin-eqn}} {{eqn | l = F_{k + 1} \phi + F_k       | r = \left({F_k + F_{k - 1} }\right) \phi + F_k       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_k \left({1 + \phi}\right) + F_{k - 1} \phi       | c =  }} {{eqn | r = F_k \phi^2 + F_{k - 1} \phi       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \phi \left({F_k \phi + F_{k - 1} }\right)       | c =  }} {{eqn | r = \phi \left({\phi^n}\right)       | c = Induction Hypothesis }} {{eqn | r = \phi^{n + 1}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and it follows by the Principle of Mathematical Induction that: :$\forall n \in \Z_{\ge 0}: \phi^n = F_n \phi + F_{n - 1}$
The proof proceeds by induction. For all $n \in \Z_{\le 0}$, let $P \left({n}\right)$ be the proposition: :$\phi^n = F_n \phi + F_{n - 1}$ This can equivalently be expressed as: For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$\phi^{-n} = F_{-n} \phi + F_{-n - 1}$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_0 \times \phi + F_{-1}       | r = F_0 \times \phi + \left({-1}\right)^0 F_1       | c = Fibonacci Number with Negative Index }} {{eqn | r = F_0 \times \phi + 1       | c = {{Defof|Fibonacci Number}} $F_1 = 1$ }} {{eqn | r = 0 \times \phi + 1       | c = {{Defof|Fibonacci Number}} $F_0 = 0$ }} {{eqn | r = 1       | c =  }} {{eqn | r = \phi^0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. ==== Basis for the Induction ==== $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{-1} \times \phi + F_{-2}        | r = \left({-1}\right)^0 F_1 \times \phi + \left({-1}\right)^{-1} F_2       | c = Fibonacci Number with Negative Index }} {{eqn | r = 1 \times \phi - F_2       | c = {{Defof|Fibonacci Number}} $F_1 = 1$ }} {{eqn | r = \phi - 1       | c = {{Defof|Fibonacci Number}}: $F_2 = 1$ }} {{eqn | r = \dfrac 1 \phi       | c = {{Defof|Golden Mean|index = 3}} }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. ==== Induction Hypothesis ==== Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\phi^{-k} = F_{-k} \phi + F_{-k - 1}$ from which it is to be shown that: :$\phi^{-\left({k + 1}\right)} = F_{-\left({k + 1}\right)} \phi + F_{-\left({k + 1}\right) - 1}$ ==== Induction Step ==== This is the induction step: {{begin-eqn}} {{eqn | l = F_{-\left({k + 1}\right)} \phi + F_{-\left({k + 1}\right) - 1}       | r = \left({-1}\right)^{-\left({k + 1}\right) + 1} F_{k + 1} \phi + \left({-1}\right)^{-\left({k + 1}\right)} F_{\left({k + 1}\right) + 1}       | c =  }} {{eqn | r = \left({-1}\right)^{k + 2} F_{k + 1} \phi - \left({-1}\right)^{k + 2} F_{k + 2}       | c =  }} {{eqn | r = \left({-1}\right)^{k + 2} \left({F_{k + 1} \phi - \left({F_{k + 1} + F_k}\right)}\right)       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \left({-1}\right)^{k + 2} \left({F_{k + 1} \left({\phi - 1}\right) - F_k}\right)       | c =  }} {{eqn | r = \left({-1}\right)^{k + 2} \left({F_{k + 1} \phi^{-1} - F_k}\right)       | c = {{Defof|Golden Mean|index = 3}} }} {{eqn | r = \left({-1}\right)^{k + 2} F_{k + 1} \phi^{-1} + \left({-1}\right)^{k + 1} F_k       | c =  }} {{eqn | r = \frac 1 \phi \left({F_{-k - 1}  + F_{-k} \phi}\right)       | c = Fibonacci Number with Negative Index }} {{eqn | r = \frac 1 \phi \left({\phi^{-k} }\right)       | c = Induction Hypothesis }} {{eqn | r = \phi^{-\left({k + 1}\right)}       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and it follows by the Principle of Mathematical Induction that: :$\forall n \in \Z_{\le 0}: \phi^n = F_n \phi + F_{n - 1}$
{{begin-eqn}} {{eqn | l = F_n \hat \phi + F_{n - 1}       | r = F_n \left({-\dfrac 1 \phi}\right) + F_{n - 1}       | c = Reciprocal Form of One Minus Golden Mean }} {{eqn | r = -\dfrac 1 \phi \left({F_n - \phi F_{n - 1} }\right)       | c =  }} {{eqn | r = -\dfrac 1 \phi \left({\left({-1}\right)^{n + 1} F_{-n} - \phi \left({-1}\right)^n F_{-\left({n - 1}\right)} }\right)       | c = Fibonacci Number with Negative Index }} {{eqn | r = -\dfrac 1 \phi \left({\left({-1}\right)^{n + 1} F_{-n} + \phi \left({-1}\right)^{n + 1} F_{-\left({n - 1}\right)} }\right)       | c =  }} {{eqn | r = \left({-1}\right)^{n + 1} \left({-\dfrac 1 \phi \left({F_{-n} + \phi F_{-n + 1} }\right)}\right)       | c =  }} {{eqn | r = \left({-1}\right)^{n + 1} \left({-\dfrac 1 \phi }\right) \left({\phi^{-n + 1} }\right)       | c = Fibonacci Number by Golden Mean plus Fibonacci Number of Index One Less }} {{eqn | r = \left({-1}\right)^n \left({\dfrac 1 {\phi^n} }\right)       | c =  }} {{eqn | r = \left({-\dfrac 1 \phi}\right)^n       | c =  }} {{eqn | r = \hat \phi^n       | c =  }} {{end-eqn}} {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$a_n = F_{n - 1} r + F_n s$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{-1} r + F_0 s       | r = \left({-1}\right)^0 F_1 r + F_0 s       | c = Fibonacci Number with Negative Index }} {{eqn | r = 1 \times r + 0 \times s       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{eqn | r = r       | c =  }} {{eqn | r = a_0       | c = {{Defof|General Fibonacci Sequence}} }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_0 r + F_1 s       | r = 0 \times r + 1 \times s       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{eqn | r = s       | c =  }} {{eqn | r = a_1       | c = {{Defof|General Fibonacci Sequence}} }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$a_k = F_{k - 1} r + F_k s$ from which it is to be shown that: :$a_{k + 1} = F_k r + F_{k + 1} s$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = a_{k + 1}       | r = a_{k - 1} + a_k       | c =  }} {{eqn | r = \left({F_{k - 2} r + F_{k - 1} s}\right) + \left({F_{k - 1} r + F_k s}\right)       | c = Induction Hypothesis }} {{eqn | r = \left({F_{k - 2} + F_{k - 1} }\right) r  + \left({F_{k - 1} + F_k}\right) s       | c =  }} {{eqn | r = F_k r + F_{k + 1} s       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: a_n = F_{n - 1} r + F_n s$ {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$b_n = c F_{n - 1} + \left({c + 1}\right) F_n - c$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = c F_{-1} + \left({c + 1}\right) F_0 - c       | r = \left({-1}\right)^0 c F_1 + \left({c + 1}\right) F_0 - c       | c = Fibonacci Number with Negative Index }} {{eqn | r = c \times 1 + \left({c + 1}\right) \times 0 - c       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{eqn | r = 0       | c =  }} {{eqn | r = b_0       | c = by definition }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = c F_0 + \left({c + 1}\right) F_1 - c       | r = c \times 0 + \left({c + 1}\right) \times 1 - c       | c = {{Defof|Fibonacci Number}}: $F_0 = 0, F_1 = 1$ }} {{eqn | r = 1       | c =  }} {{eqn | r = a_1       | c = by definition }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$b_k = c F_{k - 1} + \left({c + 1}\right) F_k - c$ from which it is to be shown that: :$b_{k + 1} = c F_k + \left({c + 1}\right) F_{k + 1} - c$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = b_{k + 1}       | r = b_{k - 1} + b_k + c       | c =  }} {{eqn | r = \left({c F_{k - 2} + \left({c + 1}\right) F_{k - 1} - c}\right) + \left({c F_{k - 1} + \left({c + 1}\right) F_k - c}\right) + c       | c = Induction Hypothesis }} {{eqn | r = c \left({F_{k - 2} + F_{k - 1} }\right) + \left({c + 1}\right) \left({F_{k - 1} + F_k}\right) - 2 c + c       | c =  }} {{eqn | r = c F_k + \left({c + 1}\right) F_{k + 1} - c       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: b_n = c F_{n - 1} + \left({c + 1}\right) F_n - c$ {{qed}}
=== Lemma === {{:Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma}} Hence also: :$b_n = F_n + \displaystyle \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} g \left({k}\right)$ Thus: {{begin-eqn}} {{eqn | l = c_n       | r = F_n + \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} \left({x f \left({k}\right) + y g \left({k}\right)}\right)       | c =  }} {{eqn | r = F_n + x \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} f \left({k}\right) + y \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} g \left({k}\right)       | c =  }} {{eqn | r = F_n + x \left({a_n - F_n}\right) + y \left({b_n - F_n}\right)       | c =  }} {{eqn | r = x a_n + y b_n + \left({1 - x - y}\right) F_n       | c =  }} {{end-eqn}} {{qed}}
From Fibonacci Number in terms of Smaller Fibonacci Numbers: :$F_{n + i} = F_n F_{i - 1} + F_{n + 1} F_i$ :$F_{n + j} = F_n F_{j - 1} + F_{n + 1} F_j$ :$F_{n + i + j} = F_{i - 1} F_{n + j} + F_i F_{n + j + 1}$ Therefore: {{begin-eqn}} {{eqn | o =       | r = F_{n + i} F_{n + j} - F_n F_{n + i + j} }} {{eqn | r = \paren {F_n F_{i - 1} + F_{n + 1} F_i} F_{n + j} - F_n \paren {F_{i - 1} F_{n + j} + F_i F_{n + j + 1} }       | c = from above }} {{eqn | r = \paren {F_n F_{i - 1} + F_{n + 1} F_i} F_{n + j} - F_n F_{i - 1} F_{n + j} - F_n F_i F_{n + j + 1}       | c =  }} {{eqn | r = \paren {F_n F_{i - 1} + F_{n + 1} F_i - F_n F_{i - 1} } F_{n + j} - F_n F_i F_{n + j + 1}       | c =  }} {{eqn | r = \paren {F_{n + 1} F_i} F_{n + j} - F_n F_i F_{n + j + 1}       | c =  }} {{eqn | r = F_i \paren {F_{n + 1} F_{n + j} - F_n F_{n + j + 1} }       | c =  }} {{eqn | r = F_i \paren {-1}^{2 n + 1} \paren {F_n F_{n + j + 1} - F_{n + 1} F_{n + j} }       | c =  }} {{eqn | r = F_i \paren {-1}^{n - j - 1} \paren {\paren {-1}^{n + j} F_n F_{n + j + 1} - \paren {-1}^{n + j} F_{n + 1} F_{n + j} }       | c =  }} {{eqn | r = F_i \paren {-1}^{n - j - 1} \paren {\paren {-1}^{n + j} F_n F_{n + j + 1} + \paren {-1}^{n + j + 1} F_{n + 1} F_{n + j} }       | c =  }} {{eqn | r = F_i \paren {-1}^{n - j - 1} F_{\paren {n + 1} - \paren {n + j + 1} }       | c = Fibonacci Number in terms of Larger Fibonacci Numbers }} {{eqn | r = F_i \paren {-1}^{n - j - 1} F_{-j}       | c =  }} {{eqn | r = F_i \paren {-1}^{n - j - 1} \paren {-1}^{j + 1} F_j       | c = Fibonacci Number with Negative Index }} {{eqn | r = \paren {-1}^n F_i F_j       | c =  }} {{end-eqn}} {{qed}} {{Namedfor|Steven Vajda}} Category:Vajda's Identity 37ahxk26iggsbeymknrp1vcf7p99e2f