Arithmetic sequence: h_{0}, h_{0} + q, h_{0} + 2q, ...

h_{n} = h_{n−1} + q = h_{0} + nq, n ≥ 0

Geometric sequence: h_{0}, qh_{0}, q^{2}h_{0}, ...

h_{n} = qh_{n−1} = q^{n}h_{0}, n ≥ 0

Ex: 1, 2, 2^{2}, ...

h_{n} = 2^{n} = number of combinations of an n-element set.

Partial sums: s_{n} = Σ_{k=0}^{n}h_{k}

Arithmetic sequence: s_{n} = Σ_{k=0}^{n}h_{0} +kq = (n+1)h_{0} + \frac{qn(n+1)}{2}

Geometric sequence: s_{n} = Σ_{k=0}^{n}q^{k}h_{0} = \{ \begin{matrix}\frac{q^{n+1}−1}{q−1}h_{0} & q ≠ 1 \\ (n + 1)h_{0} & q = 1 \end{matrix}

Fibonacci:

Suppose a pair of rabbits of the opposite sex give birth to a pair of rabbits of opposite sex every month starting with their second month. If we begin with a pair of newly born rabbits, how many rabbits are there after one year.

Let f_{n} = number of rabbits at the beginning of month n = end of month n − 1.

f_{0} =  f_{1} =  f_{2} =  f_{3} =  f_{4} =  f_{5} =

Hence f_{n} = where Lemma: s_{n} = Σ_{k=0}^{n}f_{n} = f_{n−2} − 1 Proof by induction on n. Lemma: f_{n} is even iff 3|n. Proof by induction on n. Note that f_{0} = 0 is even, f_{1} = 1 is odd, and f_{2} = 1 is odd. Suppose f_{3n} is even, f_{3n+1} is odd, and f_{3n+2} is odd. Then f_{3n+3} = f_{3n+2} + f_{3n+1}. Since odd + odd is even, f_{3n+3} is even. Then f_{3n+4} = f_{3n+3} + f_{3n+2}. Since even + odd is odd, f_{3n+4} is odd. Then f_{3n+5} = f_{3n+4} + f_{3n+3}. Since odd + even is odd, f_{3n+5} is odd.

Thm 7.1.2: f_{n} = Σ_{k=0}^{n} ( \begin{matrix}n − 1 − k \\ k \end{matrix} )

Proof: Check if g(n) = Σ_{k=0}^{n} ( \begin{matrix}n − 1 − k \\ k \end{matrix} ) satisfies g(n) = g(n − 1) + g(n − 2) and g(0) = 0 and g(1) = 1

Thm 7.1.1: f_{n} = \frac{1}{\sqrt{5}}(\frac{1+\sqrt{5}}{2})^{n} − \frac{1}{\sqrt{5}}(\frac{1−\sqrt{5}}{2})^{n}