Two ways to define a sequence.
| Kind | Looks like | To find a₅₀… |
|---|---|---|
| explicit | aₙ = 3n + 2 | plug in n = 50, done |
| recursive | a₁ = 2, aₙ₊₁ = 2aₙ − 1 | walk there — or find the pattern |
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Recursive Sequences: Each Term from the Last
A recursion hands you a starting value and a rule for the next step. Crank the handle carefully, one term at a time — and watch the index.
The problem
A sequence is defined by a₁ = 2 and aₙ₊₁ = 2aₙ − 1. What is a₅?
Crank: 2 → 3 → 5 → 9 → 17. Four applications of the rule reach the FIFTH term, because a₁ was given, not built. D applies the rule five times — the a₆ answer to an a₅ question, the off-by-one this topic exists to test.
Crank the handle — and label every term.
a₁ = 2
given, costs nothing
a₂ = 2(2) − 1 = 3, a₃ = 2(3) − 1 = 5
each term eats the previous one
a₄ = 9, a₅ = 17
four rule-applications for the fifth term
Write the labels a₁, a₂, a₃… next to the values as you go. The label column is what protects you from the off-by-one.
For far-away terms, hunt the pattern.
2, 3, 5, 9, 17, … — each is 1 more than 1, 2, 4, 8, 16
powers of two, shifted
conjecture: aₙ = 2ⁿ⁻¹ + 1
test it: n = 5 → 16 + 1 = 17 ✓
Cranking to a₂₀ by hand invites arithmetic slips; a verified explicit formula gets there in one plug. Always test the conjecture on a term you already computed before trusting it.
Two-seed recursions: the Fibonacci pattern.
a₁ = 1, a₂ = 1, aₙ = aₙ₋₁ + aₙ₋₂
each term needs TWO predecessors
1, 1, 2, 3, 5, 8, 13, …
two seeds given, then the rule takes over
When the jump depends on n.
The jump is no longer constant: going from aₙ to aₙ₊₁ adds n.
a₂ = 1 + 1 = 2, a₃ = 2 + 2 = 4
n = 1, then n = 2
a₄ = 7, a₅ = 11, a₆ = 16, a₇ = 22
jumps 3, 4, 5, 6
a₈ = 22 + 7 = 29
the last jump is 7
Shortcut for far terms: a₈ = a₁ + (1 + 2 + … + 7) = 1 + 28 = 29 ✓. Reaching a₈ takes seven jumps, n = 1 to 7.
One jump too many
Adding the jumps 1 to 8 gives 37, which is a₉. Term n needs only n − 1 jumps.
Variations: running the rule backwards, a negative multiplier, a fraction rule.
Backwards. aₙ₊₁ = 3aₙ − 2 and a₄ = 55. Find a₁.
aₙ = (aₙ₊₁ + 2) ÷ 3
undo the rule: add 2, then divide by 3
a₃ = 57 ÷ 3 = 19, a₂ = 21 ÷ 3 = 7
one step back at a time
a₁ = 9 ÷ 3 = 3
forward: 3, 7, 19, 55 ✓
What is different: the unknown is the seed. Undo the operations in reverse order: the − 2 first, then the × 3.
A negative multiplier. a₁ = 1 and aₙ₊₁ = 1 − 2aₙ. Find a₅.
a₂ = 1 − 2(1) = −1
brackets for every substitution
a₃ = 1 − 2(−1) = 3, a₄ = 1 − 2(3) = −5
minus times minus is plus
a₅ = 1 − 2(−5) = 11
the signs alternate
What is different: the terms zig-zag in sign, so every negative value goes back into the rule inside brackets.
A fraction rule. a₁ = 8 and aₙ₊₁ = aₙ ÷ 2 + 3. Find a₄.
a₂ = 4 + 3 = 7
half of 8, plus 3
a₃ = 72 + 3 = 132
keep the fraction
a₄ = 134 + 3 = 254
6.25
What is different: the terms close in on a number instead of growing. Setting L = L ÷ 2 + 3 gives L = 6, the value they approach: 8, 7, 6.5, 6.25, …
Rules that loop: find the cycle.
a₂ = 1 − 13 = 23
a₃ = 1 − 32 = −12
1 ÷ (23) = 32
a₄ = 1 − (−2) = 3 = a₁
back to the start: the cycle is 3 long
50 = 3 × 16 + 2
remainder 2
a₅₀ = a₂ = 23
the 2nd place in the cycle
Sums use the cycle too: each block of three adds to 3 + 23 − 12 = 196, so the first 9 terms add to 3 × 196 = 192.
Remainder 0
A remainder of 0 means the last place in the cycle, not the first: 48 = 3 × 16, so a₄₈ = a₃ = −12.
On the exam
Recursion items ask you to crank a few terms, find a far term, or match a rule to a list. For a far term, list terms until a formula or a repeating cycle shows, then use it — never crank to a₁₀₀. Ten-second check: put your answer back into the rule and make sure it produces the next term.
Check: run the rule once more from the answer.
forward
2(17) − 1 = 33 = a₆
consistent with the chain ✓
formula
2⁴ + 1 = 17
the conjecture agrees ✓
Now you try
Practice problem
The problem
a₁ = 1, a₂ = 1 and aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. What is a₇?
Label as you go: 1, 1, 2, 3, 5, 8, 13 — a₇ is the seventh label, 13. B overshoots one step to a₈ and C stops one early at a₆: with the terms labelled on paper, neither slip survives.