Exercises 8.4 Exercises
1.
Prove by induction that \(\displaystyle \sum_{i=0}^n (4i+1) = 2n^2 + 3n + 1\) for all integers \(n \ge 0\text{.}\)
Hint.Hint: \(4+1 = 2+3\text{.}\)
2.
Prove by induction for all integers \(n \ge 1\text{,}\) that \(\displaystyle \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\text{.}\)
Hint.Hint: The “factoring trick” will help here.
3.
Prove by induction for all integers \(n \ge 1\text{,}\) that \(\displaystyle \sum_{i=1}^n i^2 = \dfrac{n(n+1)(2n+1)}{6}\text{.}\)
Hint.Hint: The “factoring trick” will help here.
4.
Prove by induction that as long as \(r \neq 1\) and for all integers \(n \ge 0\text{,}\) that \(\displaystyle \sum_{i=0}^n r^i = \dfrac{r^{n+1}-1}{r-1}\text{.}\)
Hint.Hint: Just treat \(r\) like you would any number (that isn't equal to \(1\)— do you see why?). This is known as the formula for a finite geometric series.
5.
Using induction, prove \(\displaystyle \sum_{i=1}^n \frac{1}{i(i+1)} = \frac{n}{n+1}\) for all \(n \ge 1\text{.}\)
6.
Prove by induction that for all integers \(n \ge 6\text{,}\) \(n^2+5n+1 \le 2n^2\text{.}\)
7.
Prove by induction that \(n^2+7n+3 \le 3n^2\) for all integers \(n \ge 4\text{.}\)
8.
Prove using induction that \(n^2+3n+10 \le 4n^2\) for all \(n \ge 3\text{.}\)
9.
Prove by induction that \(2n+1 \le 2^n\) for all \(n \ge 4\text{.}\)
Hint.Hint: \(2^{k+1} = 2^k + 2^k\text{.}\)
10.
Prove by induction that \(n^2 \le 2^n\) for all \(n \ge 4\text{.}\)
Hint.Hint: Use the result from the preceding exercise.
11.
Prove by induction that \(n^3+3n+1 \le 2n^3\) for all integers \(n \ge 2\text{.}\)
Hint.Hint: \((k+1)^3 = k^3+3k^2+3k+1\text{.}\)
12.
Prove by induction that a set with \(n\) elements has \(2^n\) subsets for all \(n \ge 0\text{.}\)
Solution.Solution: Consider a set with no elements, i.e. \(\varnothing\text{.}\) The sole subset of \(\varnothing\) is \(\{\varnothing\}\text{,}\) so the set with \(n=0\) elements has \(2^0 = 1\) subset.
Inductively, assume that any set with \(k\) elements has \(2^k\) subsets if \(k\) is an integer where \(k \ge 0\text{.}\)
Now, consider the set \(X\) with \(k+1\) elements. Without loss of generality, list them:
Observe that \(X\) can be decomposed into two sets:
where
Any subset of \(X\) can be grouped into two categories, one that contains \(x'\) and one that does not. Any subset that does not contain \(x'\) is a subset of \(S\text{,}\) and we know by the inductive hypothesis that \(S\) has \(2^k\) subsets. Any subset that does contain \(x'\) can be decomposed into \(E \cup \{x'\}\) where \(E \subseteq S\text{;}\) there are also \(2^k\) of these.
So, there are \(2^k + 2^k = 2^{k+1}\) subsets of \(X\text{.}\) By induction, a set with \(n\) elements has \(2^n\) subsets.