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With the standard Fibonacci numbers $F_0=0$, $F_1=1$, this identity holds for $n\ge 0$, $k\ge 1$. At $k=1$ it is immediate. If it holds for $k$ at every $n$, apply it at $n+1$ and use $F_{n+2}=F_{n+1}+F_n$ to obtain $F_{n+k+1}=F_{k+1}{F}_{n+1}+F_k{F}_n$, proving the assertion by mathematical induction. Taking $k=n$ gives $F_{2n}=F_n(F_{n+1}+F_{n-1})$ for $n\ge 1$, linking the doubling formula to the Lucas numbers.

For $\phi =(1+\sqrt{5})/2$ and $\psi =(1-\sqrt{5})/2$, each power sequence satisfies the recurrence relation $x_{n+2}=x_{n+1}+x_n$, because both bases solve $t^2=t+1$. Their sum starts at $2,1$, so uniqueness of a recursively defined sequence proves that it is the Lucas number sequence. Alternatively, diagonalize the companion matrix $\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ and use $(1,2{)}^T=(\phi ,1{)}^T+(\psi ,1{)}^T$. The formula explains the exponential growth governed by the golden ratio and the alternating decaying correction ${\psi }^n$.

The Lucas numbers obey the same second-order recurrence as the Fibonacci numbers but start with $2,1$, so their sequence is $2,1,3,4,7,11,\dots$. For $n\ge 1$, $L_n=F_{n-1}+F_{n+1}$. Substitution in the Fibonacci addition formula gives $F_{2n}=F_n{L}_n$. Consecutive Fibonacci numbers are coprime by the Euclidean algorithm; since $L_n=F_n+2F_{n-1}$, Bézout's identity gives $\gcd (F_n,L_n)=\gcd (F_n,2)\le 2$.

For the Fibonacci addition formula, use mathematical induction on $k$, with the assertion applying to every $n\ge 1$. At $k=2$, $F_{n+2}=F_{n+1}+F_n=F_2{F}_{n+1}+F_1{F}_n$. If the formula holds for some $k\ge 2$ and all $n$, apply it at $n+1$ and use the Fibonacci number recurrence:This is the assertion at $k+1$, completing the mathematical induction. ThusTaking $k=n\ge 2$ gives the doubling identity

For the Lucas numbers, $L_2=F_3+F_1=3$, and $L_3=F_4+F_2=4=L_2+L_1$. For $n\ge 4$, both expressions for earlier Lucas numbers are available, andHence $\boxed{L_1=1,L_2=3,L_n=L_{n-1}+L_{n-2}(n\ge 3)}$. The separate $n=3$ check matters because the defining expression for $L_1$ was given separately.

The Euclidean algorithm and the Fibonacci number recurrence give, for $n\ge 2$,Repeating reduces to $\gcd (F_2,F_1)=1$; the case $n=1$ is immediate. Therefore $\boxed{\gcd (F_n,F_{n+1})=1(n\ge 1)}$.

For $n\ge 2$, $L_n=F_n+2F_{n-1}$, so a common divisor $d$ of $F_n,L_n$ divides both $F_n$ and $2F_{n-1}$. Since $F_n,F_{n-1}$ are coprime, Bézout's identity gives integers $s,t$ with $s{F}_n+t{F}_{n-1}=1$. Multiplying by $2$ shows $d\mid 2$. Conversely any common divisor of $F_n$ and $2$ divides $L_n$. Including $n=1$, this proves the stronger greatest common divisor identity

Adding the Fibonacci recurrence and the identical recurrence relation for the Lucas numbers givesLinearity preserves this linear recurrence relation; the initial values come from $0+2$ and $1+1$.

The update must place the sum of the two current entries in the first coordinate and retain the old first coordinate in the second. Thus choose the matrixBy matrix multiplication, $A(L_n,L_{n-1}{)}^T=(L_n+L_{n-1},L_n{)}^T=(L_{n+1},L_n{)}^T$ for every $n\ge 1$, using the given recurrence relation. This is the companion-matrix representation of the Lucas numbers.

Use mathematical induction on $n$. For $n=0$, the identity matrix gives $(1,2{)}^T=(L_1,L_0{)}^T$. If $A^n(1,2{)}^T=(L_{n+1},L_n{)}^T$, applying $A$ and using part (a) gives $A^{n+1}(1,2{)}^T=(L_{n+2},L_{n+1}{)}^T$. ThereforeThe base case is essential: the starting vector represents consecutive Lucas numbers, not two arbitrary initial values.