Using the binomial theorem

The theorem is not only for algebra. It turns hard arithmetic, comparisons and divisibility proofs into a few easy terms.

Big powers by hand. Split the base into a round number and a small one. For 99⁴, write 99 = 100 – 1 and use the row 1, 4, 6, 4, 1 with alternating signs:

• (100 – 1)⁴ = 100⁴ – 4·100³ + 6·100² – 4·100 + 1
• = 100000000 – 4000000 + 60000 – 400 + 1
• = 96059601. (Check: 99² = 9801, and 9801² = 96059601.)

Add the positive terms together first and the negative ones together, then subtract once: there are fewer chances to slip that way.

Which is larger? Is (1.02)⁵⁰⁰ bigger than 10? Write it as (1 + 0.02)⁵⁰⁰ and look at just the first two terms:

• (1 + 0.02)⁵⁰⁰ = 1 + 500 × 0.02 + (other positive terms) = 1 + 10 + ... = 11 + ...
• That is already more than 10, and the rest only adds to it, so (1.02)⁵⁰⁰ > 10.

Divisibility. To say a number is divisible by 16, show it equals 16 × (a whole number). Show that 5ⁿ – 4n – 1 is divisible by 16:

• 5ⁿ = (1 + 4)ⁿ = 1 + ⁿC₁·4 + ⁿC₂·4² + ⁿC₃·4³ + ...
• = 1 + 4n + 16(ⁿC₂ + ⁿC₃·4 + ...).
• Move 1 + 4n across: 5ⁿ – 4n – 1 = 16 × (a whole number). Done. (n = 2: 25 – 8 – 1 = 16.)

The same idea gives remainders: 8ⁿ = (7 + 1)ⁿ = 7 × (whole number) + 1, so 8ⁿ always leaves remainder 1 when divided by 7.

Adding and subtracting twin expansions. (a + b)ⁿ and (a – b)ⁿ have the same terms with alternating signs. Adding them doubles the terms with even powers of b and cancels the rest; subtracting keeps only the odd-power terms, doubled.

• (√3 + 1)⁴ + (√3 – 1)⁴ = 2[(√3)⁴ + 6(√3)²·1² + 1⁴]
• = 2[9 + 18 + 1] = 56, a whole number although each bracket has a surd.

Sums with Σ. Σ means 'add up'. Σ (r = 0 to n) 2ʳ ⁿCᵣ means put r = 0, 1, 2, ..., n into 2ʳ ⁿCᵣ and add: ⁿC₀ + 2·ⁿC₁ + 2²·ⁿC₂ + ... + 2ⁿ·ⁿCₙ. That is exactly the expansion of (1 + 2)ⁿ, so the sum is 3ⁿ.