Binomial Theorem

Pascal's triangle is a triangular arrangement of numbers in which the first and last entry of every row is 1, and every other entry is the sum of the two entries just above it in the previous row. Row n (counting rows from n = 0) of the triangle contains exactly the numbers that appear as coefficients when (a + b)ⁿ is expanded.

For example, expanding (a+b)⁰, (a+b)¹, (a+b)², (a+b)³, (a+b)⁴ gives the coefficient rows 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1. These numbers are exactly the binomial coefficients ⁿC_r = n! / (r!(n−r)!), read across row n as ⁿC₀, ⁿC₁, …, ⁿC_n.

Worked example. Write row 5 of Pascal's triangle and use it to expand (a+b)⁵. Row 5 is built by adding adjacent entries of row 4 (1 4 6 4 1): the ends stay 1, and 1+4=5, 4+6=10, 6+4=10, 4+1=5. So row 5 is 1 5 10 10 5 1, giving (a+b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵.

This row-building rule is exactly the identity ⁿC_r = ⁿ⁻¹C_r−1 + ⁿ⁻¹C_r, called Pascal's rule, which lets each new row be built from the previous one without recomputing factorials.

The binomial theorem states that for any real numbers a, b and any positive integer n, (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹b + ⁿC₂ aⁿ⁻²b² + … + ⁿC_n bⁿ, that is, (a+b)ⁿ = Σ_r=0^{n n}C_r a^n−r b^r.

Three patterns are worth noting: the expansion always has exactly (n+1) terms; the power of a decreases step by step from n to 0 while the power of b increases from 0 to n, and in every term the two exponents always add up to n; the coefficients are precisely the binomial coefficients ⁿC_r, which can be read straight off Pascal's triangle.

Worked example. Expand (2x − 3)⁵ using the binomial theorem. Here a = 2x, b = −3, n = 5, so (2x−3)⁵ = ⁵C₀(2x)⁵ + ⁵C₁(2x)⁴(−3) + ⁵C₂(2x)³(−3)² + ⁵C₃(2x)²(−3)³ + ⁵C₄(2x)(−3)⁴ + ⁵C₅(−3)⁵.

= 32x⁵ + 5(16x⁴)(−3) + 10(8x³)(9) + 10(4x²)(−27) + 5(2x)(81) + (−243)

= 32x⁵ − 240x⁴ + 720x³ − 1080x² + 810x − 243. As a check, putting x = 1 gives 32−240+720−1080+810−243 = −1, matching (2−3)⁵ = (−1)⁵ = −1.

In (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹b + … + ⁿC_n bⁿ, the term in position (r+1), counting the first term as T₁, is called the general term: T_r+1 = ⁿC_r a^n−r b^r. It is called "general" because giving r the values 0, 1, 2, …, n in turn produces every term of the expansion, so it is the key tool for finding one particular term (a specific coefficient, a term with a given power of x, a term independent of x) without expanding the whole binomial.

Worked example. Find the general term of (x + 2y)⁷, and use it to write the 5th term. Here a = x, b = 2y, n = 7, so T_r+1 = ⁷C_r x^7−r(2y)^r. The 5th term corresponds to r + 1 = 5, that is r = 4: T₅ = ⁷C₄ x³(2y)⁴ = 35 · x³ · 16y⁴ = 560 x³y⁴.

When the binomial has a minus sign, as in (a − b)ⁿ, the general term carries an extra factor (−1)^r: T_r+1 = (−1)^{r n}C_r a^n−r b^r.

The expansion of (a+b)ⁿ has (n+1) terms in total. Whether this count is odd or even decides how many "middle" terms the expansion has.

Case 1: n is even. Then (n+1) is odd, so there is exactly one middle term, and it is the ((n/2)+1)th term, i.e., T_(n/2)+1.

Worked example (n even). Find the middle term in the expansion of (x/3 + 9y)¹⁰. Here n = 10 is even, so the middle term is the (10/2 + 1) = 6th term, i.e., r = 5: T₆ = ¹⁰C₅ (x/3)⁵(9y)⁵ = 252 · (x⁵/243) · 59049y⁵ = 252 × 243 · x⁵y⁵ = 61236 x⁵y⁵, since 59049/243 = 243.

Case 2: n is odd. Then (n+1) is even, so there are exactly two middle terms, namely the ((n+1)/2)th and ((n+3)/2)th terms.

Worked example (n odd). Find the middle terms in the expansion of (x + 1/x)⁷. Here n = 7 is odd, so the middle terms are the 4th and 5th terms (r = 3 and r = 4): T₄ = ⁷C₃ x⁴(1/x)³ = 35x, and T₅ = ⁷C₄ x³(1/x)⁴ = 35/x.

Finding a term independent of x. To find the term that has no x in it at all, write the general term T_r+1, collect the power of x into a single exponent, set that exponent equal to 0, and solve for r.

Worked example. Find the term independent of x in the expansion of (x − 2/x²)¹⁵. The general term is T_r+1 = ¹⁵C_r x^15−r(−2/x²)^r = ¹⁵C_r(−2)^r x^15−3r. For the term independent of x, 15 − 3r = 0, so r = 5. Then T₆ = ¹⁵C₅(−2)⁵ = 3003 × (−32) = −96096.

Computing powers of numbers close to a round number. Writing a number as (round number ± small number) and expanding lets us compute an exact power without long multiplication.

Worked example. Evaluate 99⁵ using the binomial theorem. Write 99 = 100 − 1, so 99⁵ = (100−1)⁵ = Σ ⁵C_r 100^5−r(−1)^r = 100⁵ − 5(100⁴) + 10(100³) − 10(100²) + 5(100) − 1 = 10,000,000,000 − 500,000,000 + 10,000,000 − 100,000 + 500 − 1 = 9,509,900,499.

Putting particular values of x into (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿC_nxⁿ produces useful identities among the binomial coefficients.

Putting x = 1: ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿC_n = 2ⁿ, i.e., the sum of all binomial coefficients in row n is 2ⁿ.

Putting x = −1 (for n ≥ 1): ⁿC₀ − ⁿC₁ + ⁿC₂ − … + (−1)^{n n}C_n = 0. Combining this with the x = 1 identity shows that the sum of coefficients at even positions equals the sum of coefficients at odd positions, and each of these two sums equals 2ⁿ⁻¹.

Worked example. The sum of the binomial coefficients in the expansion of (1 + x)ⁿ is 256. Find n. By the identity above, 2ⁿ = 256 = 2⁸, so n = 8.