Prove the following proposition: \[ \left( \sum_{i=0}^n 2^i \right) = 2^{n+1} - 1 \text{ for any integer }n \ge 0 \] or in other words, \(2^0 + 2^1 + 2^2 + \ldots + 2^n = 2^{n+1} - 1\)
(Ch10, Ex #25) Let \(F_i\) be the \(i\)th element of the Fibonacci sequence. Prove that \(F_1 + F_2 + F_3 + \cdots + F_n = F_{n+2} - 1\).
Prove that any integer numbers of cents \(\ge 8\) can be made using a combination of \(5\textcent\) and \(3\textcent\) coins.
Prove that for all integers \(n \ge 1\), \(8^n – 3^n\) is divisible by 5.
For all \(n \in \mathbb{N}\), prove that \(2^n \le 2^{n+1} - 2^{n-1} - 1\).