Lesson 1 of 12 · Algebra Foundations
Arithmetic & Algebraic Manipulation
Before reaching for a theorem, look for structure. Telescoping, nested fractions, symmetric sums, and strategic substitution turn expressions that look like heavy computation into a few lines of exact arithmetic — the kind the AMC 10 expects without a calculator.
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Learning objectives
- Recognize telescoping structure in a sum or product and collapse it in one line.
- Simplify nested (continued) fractions by working from the innermost term outward.
- Rewrite symmetric expressions in two variables using their sum and product instead of solving for each variable.
- Use round-number substitution to compute large products, squares, or differences exactly and quickly.
The core idea
Direct computation is a fallback, not a first move. Every technique in this lesson does the same thing: it rewrites an expression so that most of it cancels, repeats, or simplifies — leaving only a small amount of arithmetic at the end. The skill is not any single formula; it's the habit of pausing before computing to ask "does this have exploitable structure?"
Four patterns cover most of what shows up on the AMC 10:
- Telescoping — a sum or product where consecutive terms cancel, leaving only the first and last pieces.
- Nested fractions — continued fractions that must be resolved from the inside out.
- Symmetric sums — expressions in two (or more) unknowns that depend only on their sum and product, not their individual values.
- Strategic substitution — replacing an ugly number with "a round number plus or minus a small correction" to make the arithmetic exact and fast.
Telescoping via partial fractions
1 / (n(n+1)) = 1/n − 1/(n+1)
Splitting a product of two linear factors in the denominator into a difference of simple fractions is the single most common telescoping setup on the AMC 10. The general form is 1/((n+a)(n+b)) = [1/(b−a)]·(1/(n+a) − 1/(n+b)).
Symmetric sums from sum and product
a² + b² = (a + b)² − 2ab · a³ + b³ = (a + b)³ − 3ab(a + b)
If a problem gives you a + b and ab (or lets you find
them easily), you almost never need a and b
individually. These identities are previewed here and formalized fully in
the upcoming "Roots, Coefficients & Vieta's Formulas" lesson.
Worked example 1 — Telescoping sum
Problem
Evaluate 1/(1·2) + 1/(2·3) + 1/(3·4) + 1/(4·5) + 1/(5·6).
Key insight
Each term has the form 1/(n(n+1)), which splits into 1/n − 1/(n+1).
Solution
Rewrite every term: the sum becomes (1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + (1/4 − 1/5) + (1/5 − 1/6). Every interior term cancels with its neighbor, leaving only the first and last pieces: 1 − 1/6 = 5/6.
Takeaway
You never need to find a common denominator across all five fractions — the decomposition does the cancellation for you.
Worked example 2 — Nested fraction
Problem
Simplify 1 + 1/(1 + 1/(1 + 1/2)) to a single fraction in lowest terms.
Key insight
Resolve strictly from the innermost fraction outward — there is no shortcut that skips a layer.
Solution
Innermost: 1 + 1/2 = 3/2. Next layer: 1 + 1/(3/2) = 1 + 2/3 = 5/3. Outer layer: 1 + 1/(5/3) = 1 + 3/5 = 8/5.
Takeaway
Nested fractions are mechanical, not clever — the only risk is arithmetic slips, so write out each layer explicitly rather than trying to shortcut in your head.
Worked example 3 — Symmetric-sum shortcut
Problem
If a + b = 10 and ab = 21, find a² + b² without solving for a and b individually.
Key insight
a² + b² = (a + b)² − 2ab — a direct algebraic identity, no need to factor 21 into specific roots.
Solution
(a + b)² − 2ab = 10² − 2(21) = 100 − 42 = 58.
Alternate method
Solving t² − 10t + 21 = 0 gives t = 3 or t = 7, and 3² + 7² = 9 + 49 = 58 — confirms the identity, but takes an extra step and requires the quadratic to factor nicely.
Takeaway
Whenever you see "a + b = ..., ab = ..., find (some symmetric expression)," reach for the identity before reaching for the quadratic formula.
Worked example 4 — Strategic substitution
Problem
Compute 998 × 1002 exactly, without long multiplication.
Key insight
Both factors sit near the round number 1000: 998 = 1000 − 2 and 1002 = 1000 + 2.
Solution
(1000 − 2)(1000 + 2) = 1000² − 2² = 1,000,000 − 4 = 999,996.
Takeaway
Any product or square involving numbers close to a round value is worth rewriting as (round ± small) before multiplying directly — this pattern is formalized as the difference-of-squares identity in the next lesson.
Strategy notes
Pause before you compute
On a no-calculator exam, if a problem seems to demand heavy arithmetic, that's usually a signal you're missing the intended structure — not a signal to compute faster. Spend five seconds scanning for telescoping, symmetry, or a useful substitution before starting to multiply.
Write out the cancellation explicitly the first few times
It's tempting to jump straight to "first term minus last term" for a telescoping sum. Until that's automatic, write out at least the first two and last two terms so a boundary error (an extra or missing term) is easy to catch.
Common mistakes
Off-by-one errors in telescoping bounds
When a sum runs from n = 1 to n = k, double-check exactly which boundary terms survive. It's easy to include or exclude one extra term, especially when the index doesn't start at 1 or the step size isn't 1.
Collapsing a nested fraction from the outside in
Nested fractions must be resolved innermost-first. Trying to simplify the outer expression before the inner one is fully resolved almost always produces the wrong answer.
Approximating when an exact answer is required
The AMC 10 always has an exact, calculator-free answer. If your approach leaves you estimating decimals, you're very likely missing the intended algebraic shortcut.
Practice
Type your answer and check it — 30 questions across four difficulty
tiers, up to three tries per question before the solution is shown.
Fractions and decimals both work
(e.g. 3/4 or 0.75). If your first-try accuracy is
below 70%, you'll get one reinforcement pass through the questions you missed.