Binomial theorem

Start and end every row with 1; each inside number is the sum of the two directly above it. Rows: 1 / 1 1 / 1 2 1 / 1 3 3 1.

The coefficients of (a + b)ⁿ. E.g. row 3 (1, 3, 3, 1) → (a + b)³ = a³ + 3a²b + 3ab² + b³.

Take r = 3 factors counting down from 8 on top, r! on the bottom: ⁸C₃ = (8×7×6)/(3×2×1) = 56. The big factorials cancel — never expand them in full. (ⁿCᵣ = n!/(r!(n − r)!).)

5!/(2! 3!) = (5 × 4)/(2 × 1) = 10.

Type n, then MATH → ▶ (PRB) → 3: nCr, then r, then ENTER. E.g. 10 nCr 4 = 210.

Small n (about ≤ 6) and you want the WHOLE expansion → Pascal's triangle is fastest. Large n, or you only need ONE term/coefficient → use ⁿCᵣ (the general term ⁿCᵣaⁿ⁻ʳbʳ) and skip the rest.

n + 1 terms. E.g. (x + 2)⁹ has 10 terms.

The power of a falls from n to 0; the power of b rises from 0 to n; in every term the two powers sum to n.

Both equal 1 (the first and last coefficient of every row). The row is symmetric: ⁿCᵣ = ⁿCₙ₋ᵣ.

ⁿCᵣ is the entry in row n, position r (counting from 0). Row 5 = ⁵C₀, ⁵C₁, …, ⁵C₅ = 1, 5, 10, 10, 5, 1.

Multiply each coefficient ⁿCᵣ by a falling power of a and a rising power of b: aⁿ + ⁿC₁aⁿ⁻¹b + … + bⁿ.

Coeffs 1, 4, 6, 4, 1 with rising powers of 3: x⁴ + 12x³ + 54x² + 108x + 81.

Raise the WHOLE term: (3x)² = 9x², not 3x². Always bracket the term before squaring/cubing.

Use −b as the second term; even powers come out +, odd powers −. So the signs alternate: + − + − …

1 + 4(−2x) + 6(−2x)² + 4(−2x)³ + (−2x)⁴ = 1 − 8x + 24x² − 32x³ + 16x⁴.

Take r = 0, 1, 2, 3 in turn (the lowest powers of x) and stop — no need for the whole expansion.

1 + ¹⁰C₁x + ¹⁰C₂x² = 1 + 10x + 45x².

(1 + rate)ⁿ is the compound-interest factor; expanding it gives the value (or a quick approximation) by hand.

The two powers in every term sum to n, and there are n + 1 terms in total.

Two errors — the sign and the 3. Here a = 3x (not x) and b = −2 (not 2). Correct: ⁵C₃ (3x)²(−2)³ = 10 × 9 × (−8) x² = −720x². Always raise the WHOLE bracket term — number, variable and sign — to its power. (General term: ⁿCᵣaⁿ⁻ʳbʳ.)

Use the general term ⁿCᵣ aⁿ⁻ʳ bʳ; set the exponent of x equal to the power you want, solve for r, then compute that one term.

Write the general term, find the r that gives that power of x, and compute the coefficient (raising the whole coefficient/sign to the power).

The power of x is 0. Set the exponent of x to 0, solve for r, then compute that term.

Write that coefficient via the general term, set it equal to the given value, and solve. E.g. (x+k)⁷ coeff x⁵ = 63 → 21k² = 63 → k = ±√3.

An even power of the unknown (e.g. k²) gives two values. Check for a restriction like 'k > 0' before keeping both.

Use the simplest coefficient: ⁿC₂ = n(n − 1)/2 gives a quadratic in n; solve for the positive integer.

Use ⁿC₁k = (x coefficient) and ⁿC₂k² = (x² coefficient); eliminate k and solve for n, then k.

r = 2: ⁶C₂(2x)⁴(−3)² = 15 × 16 × 9 = 2160.

Form both equations and divide one by the other to eliminate a variable.