Problem-Solving StrategyDifficulty 5-7094 tagged problems

Find a Pattern

Find a Pattern is the habit of solving small examples and extracting the structure that persists. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\text{solve } n = 1, 2, 3, \dots \;\Rightarrow\; \text{conjecture the rule, then verify it holds for all } n.$$

Work out the smallest cases, spot the structure that persists, and turn that observation into a formula or rule for the general case.

Why It Works

1
Compute the answer for $n = 1, 2, 3, \dots$ exactly, keeping the results in a form (sum, product, factored) where structure can show.
2
Compare consecutive results: look for a closed form such as $n^2$, $2^n - 1$, or a constant ratio or difference.
3
Form a conjecture for general $n$, then test it on a fresh case you have not used — agreement is strong evidence, disagreement kills the guess.
4
For a contest answer the verified pattern usually suffices; a full proof (often by induction) seals it.

Worked Examples

Example 1

Find a formula for the sum of the first $n$ odd numbers $1 + 3 + 5 + \cdots + (2n-1)$.

1
Tabulate partial sums: $1 = 1$, $1+3 = 4$, $1+3+5 = 9$, $1+3+5+7 = 16$.
2
The results $1, 4, 9, 16$ are exactly the perfect squares $1^2, 2^2, 3^2, 4^2$.
3
Conjecture that the sum of the first $n$ odd numbers equals $n^2$.
4
Check $n = 5$: $1+3+5+7+9 = 25 = 5^2$, confirming the pattern.

Answer:

$1 + 3 + 5 + \cdots + (2n-1) = n^2$.

Contest level

Find a closed form for the sum of the first $n$ cubes $1^3 + 2^3 + \cdots + n^3$.

1
Tabulate partial sums: $1 = 1$, $1+8 = 9$, $1+8+27 = 36$, $1+8+27+64 = 100$.
2
The results $1, 9, 36, 100$ are perfect squares: $1^2, 3^2, 6^2, 10^2$.
3
The bases $1, 3, 6, 10$ are the triangular numbers $\frac{n(n+1)}{2}$, so the sum looks like $\left(\frac{n(n+1)}{2}\right)^2$.
4
Check $n = 5$: the sum is $1+8+27+64+125 = 225 = 15^2$ and $\frac{5\cdot 6}{2} = 15$, confirming the conjecture.

Answer:

$1^3 + 2^3 + \cdots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$, the square of the $n$th triangular number.

Olympiad / Challenge

Find a closed form for $S_n = 1\cdot 1! + 2\cdot 2! + 3\cdot 3! + \cdots + n\cdot n!$.

1
Tabulate: $S_1 = 1$, $S_2 = 1 + 4 = 5$, $S_3 = 5 + 18 = 23$, $S_4 = 23 + 96 = 119$.
2
Compare each value to nearby factorials: $1 = 2! - 1$, $5 = 3! - 1$, $23 = 4! - 1$, $119 = 5! - 1$.
3
Conjecture $S_n = (n+1)! - 1$; the structure behind it is $k\cdot k! = (k+1)! - k!$, so the sum telescopes.
4
Telescoping confirms it: $\sum_{k=1}^n \big((k+1)! - k!\big) = (n+1)! - 1!= (n+1)! - 1$.

Answer:

$S_n = (n+1)! - 1$.

Going Deeper

Spotting the pattern is only step one; the rule it suggests ($k^3$ summing to a triangular-number square, or $k\cdot k!$ telescoping) usually has a clean algebraic reason that converts a guess into a proof.
Pattern-finding seeds nearly every AIME number-theory and sequence problem: compute small cases, conjecture, then justify by induction or telescoping — the conjecture also tells you which closed form to aim for.
Pitfall: a finite run of agreement is not proof. Sequences like the number of regions cut by chords joining $n$ points on a circle read $1, 2, 4, 8, 16$ and then jump to $31$, not $32$ — always verify a fresh case and seek the underlying reason.

Spot the Signal

Look for problems where the key step is solving small examples and extracting the structure that persists.
You can describe the hard part as solving small examples and extracting the structure that persists, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the problem-solving bottleneck that calls for find a pattern, then rewrite the givens around it.
Name the relation that makes Find a Pattern legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Find a Pattern just because the surface looks familiar; verify the required condition first.
Applying Find a Pattern because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let find a pattern reveal the strategic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is solving small examples and extracting the structure that persists.

Back to techniques

Problem-Solving StrategyDifficulty 5-7094 tagged problems

Find a Pattern

Practice this technique Match my level

The Key Result

$$\text{solve } n = 1, 2, 3, \dots \;\Rightarrow\; \text{conjecture the rule, then verify it holds for all } n.$$

Work out the smallest cases, spot the structure that persists, and turn that observation into a formula or rule for the general case.

Why It Works

1
Compute the answer for $n = 1, 2, 3, \dots$ exactly, keeping the results in a form (sum, product, factored) where structure can show.
2
Compare consecutive results: look for a closed form such as $n^2$, $2^n - 1$, or a constant ratio or difference.
3
Form a conjecture for general $n$, then test it on a fresh case you have not used — agreement is strong evidence, disagreement kills the guess.
4
For a contest answer the verified pattern usually suffices; a full proof (often by induction) seals it.

Worked Examples

Example 1

Find a formula for the sum of the first $n$ odd numbers $1 + 3 + 5 + \cdots + (2n-1)$.

1
Tabulate partial sums: $1 = 1$, $1+3 = 4$, $1+3+5 = 9$, $1+3+5+7 = 16$.
2
The results $1, 4, 9, 16$ are exactly the perfect squares $1^2, 2^2, 3^2, 4^2$.
3
Conjecture that the sum of the first $n$ odd numbers equals $n^2$.
4
Check $n = 5$: $1+3+5+7+9 = 25 = 5^2$, confirming the pattern.

Answer:

$1 + 3 + 5 + \cdots + (2n-1) = n^2$.

Contest level

Find a closed form for the sum of the first $n$ cubes $1^3 + 2^3 + \cdots + n^3$.

1
Tabulate partial sums: $1 = 1$, $1+8 = 9$, $1+8+27 = 36$, $1+8+27+64 = 100$.
2
The results $1, 9, 36, 100$ are perfect squares: $1^2, 3^2, 6^2, 10^2$.
3
The bases $1, 3, 6, 10$ are the triangular numbers $\frac{n(n+1)}{2}$, so the sum looks like $\left(\frac{n(n+1)}{2}\right)^2$.
4
Check $n = 5$: the sum is $1+8+27+64+125 = 225 = 15^2$ and $\frac{5\cdot 6}{2} = 15$, confirming the conjecture.

Answer:

$1^3 + 2^3 + \cdots + n^3 = \left(\dfrac{n(n+1)}{2}\right)^2$, the square of the $n$th triangular number.

Olympiad / Challenge

Find a closed form for $S_n = 1\cdot 1! + 2\cdot 2! + 3\cdot 3! + \cdots + n\cdot n!$.

1
Tabulate: $S_1 = 1$, $S_2 = 1 + 4 = 5$, $S_3 = 5 + 18 = 23$, $S_4 = 23 + 96 = 119$.
2
Compare each value to nearby factorials: $1 = 2! - 1$, $5 = 3! - 1$, $23 = 4! - 1$, $119 = 5! - 1$.
3
Conjecture $S_n = (n+1)! - 1$; the structure behind it is $k\cdot k! = (k+1)! - k!$, so the sum telescopes.
4
Telescoping confirms it: $\sum_{k=1}^n \big((k+1)! - k!\big) = (n+1)! - 1!= (n+1)! - 1$.

Answer:

$S_n = (n+1)! - 1$.

Going Deeper

Spotting the pattern is only step one; the rule it suggests ($k^3$ summing to a triangular-number square, or $k\cdot k!$ telescoping) usually has a clean algebraic reason that converts a guess into a proof.
Pattern-finding seeds nearly every AIME number-theory and sequence problem: compute small cases, conjecture, then justify by induction or telescoping — the conjecture also tells you which closed form to aim for.
Pitfall: a finite run of agreement is not proof. Sequences like the number of regions cut by chords joining $n$ points on a circle read $1, 2, 4, 8, 16$ and then jump to $31$, not $32$ — always verify a fresh case and seek the underlying reason.

Spot the Signal

Look for problems where the key step is solving small examples and extracting the structure that persists.
You can describe the hard part as solving small examples and extracting the structure that persists, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the problem-solving bottleneck that calls for find a pattern, then rewrite the givens around it.
Name the relation that makes Find a Pattern legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Find a Pattern just because the surface looks familiar; verify the required condition first.
Applying Find a Pattern because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let find a pattern reveal the strategic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is solving small examples and extracting the structure that persists.