Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — using a recurrence relation
Given u1=3 and un+1=2un+1, find the first four terms.
- 1u1=3 is given.A recurrence is useless without its starting value.
- 2u2=2(3)+1=7Substitute the previous term.
- 3u3=2(7)+1=15
- 4u4=2(15)+1=31Each term needs the one before it, which is the drawback of a recurrence.
- 53,7,15,31A pattern is visible: each term is one less than a power of 2, suggesting un=2n+1−1 — a conjecture that induction could then prove.
Example 2 — the limit of a sequence
Find n→∞limn+23n+1.
- 1Both top and bottom grow without bound, so the ratio is not obvious.
- 2Divide every term by n, the highest power in the denominator.
- 31+n23+n1
- 4As n→∞, both n1 and n2 tend to 0.
- 5The limit is 13=3.Check with n=1000: 10023001=2.995 ✓
Example 3 — proving a formula by induction
Given u1=3 and un+1=2un−1, prove that un=2n+1.
- 1Base case: when n=1 the formula gives 21+1=3, matching u1 ✓Always verify the smallest case explicitly.
- 2Assume uk=2k+1 for some k≥1.The inductive hypothesis.
- 3uk+1=2uk−1Use the recurrence.
- 4=2(2k+1)−1Here is where the assumption is used — essential.
- 5=2k+1+2−1=2k+1+1Which is the formula with n=k+1 ✓
- 6By induction, un=2n+1 for all n≥1.The conclusion sentence is part of the proof.