The binomial theorem

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Multiplying out $(a + b)^{n}$ lists every product $a^{n - k} b^{k}$ . It does not say how many copies of that product appear. For $n = 2$ the copies are areas you can point at. For a larger $n$ the square stops being a picture, and the count has to be named before the product is written down.

Cut one square

Two extra freezes sit on the figure. The side lengths $a$ and $b$ stay positive, so each piece has an area. The identity written underneath is claimed for every real or complex value, including the ones this square cannot draw.

Drag a or b. Both stay positive, so the figure stays a square of side a+b. The corner square of side a, the opposite corner of side b, and the two rectangles are the four terms of (a+b)^2.

The outer square has side $a + b$ . One corner is a square of side $a$ , the opposite corner is a square of side $b$ , and the two rectangles each have area $ab$ . That is the first of the three square formulas,

$(a + b)^{2} = a^{2} + 2 ab + b^{2} .$

Replace $b$ by $- b$ . The cross term changes sign, and the same identity becomes

$(a - b)^{2} = a^{2} - 2 ab + b^{2} .$

The second formula is that substitution. It is not a separate expansion. The board cannot show the minus sign: a negative length is not a side. What the substitution keeps is the algebra, after the positive square has already shown where $2 ab$ comes from.

Factor the difference of squares

The third square formula multiplies the sum by the difference instead of squaring either one:

$(a - b) (a + b) = a^{2} - b^{2} .$

The cross terms cancel. The same cancellation, repeated, is a finite geometric sum. For $n = 1, 2, \dots$ and every real or complex $a, b$ with $a \neq = b$ ,

$\sum_{k = 0}^{n} a^{n - k} b^{k} = a^{n} + a^{n - 1} b + \dots + a b^{n - 1} + b^{n} = \frac{a ^{n + 1} - b ^{n + 1}}{a - b} .$

Each term is the previous term times $b / a$ , so this is the closed quotient on finite geometric and arithmetic series, written with first term $a^{n}$ and ratio $b / a$ . The hypothesis $a \neq = b$ is the hypothesis $q \neq = 1$ on that page. Set $a = b$ and the denominator is zero. Every term is then $a^{n}$ , there are $n + 1$ of them, and the sum is $(n + 1) a^{n}$ . The quotient has stopped; the row has not.

The share of each product

The count that the square was showing for $n = 2$ has a name that still works when the exponent is not a positive integer. For $k = 1, 2, \dots$ and every real $α$ ,

$(k α) := \frac{α}{1} \cdot \frac{α - 1}{2} \cdot \frac{α - 2}{3} \dots \frac{α - k + 1}{k},$

and

$(0 α) := 1.$

The first two numerical checks are $(2 3) = (3 \cdot 2) / (1 \cdot 2) = 3$ and $(3 5) = (5 \cdot 4 \cdot 3) / (1 \cdot 2 \cdot 3) = 10$ .

When $n$ is a positive integer, $(k n)$ is the number of ways to choose which $k$ of the $n$ factors contribute $b$ , the rest contributing $a$ . That is the debt the product was defined to pay. For $n = 1, 2, \dots$ and every real or complex $a, b$ ,

$(a + b)^{n} = a^{n} + (1 n) a^{n - 1} b + (2 n) a^{n - 2} b^{2} + \dots + (n - 1 n) a b^{n - 1} + b^{n} .$

With the sum sign, the same expansion is

$(a + b)^{n} = \sum_{k = 0}^{n} (k n) a^{n - k} b^{k} .$

These are (0.16) and (0.17). Substitute $- b$ for $b$ in (0.17) and the second general formula is immediate:

$(a - b)^{n} = \sum_{k = 0}^{n} (k n) (- 1)^{k} a^{n - k} b^{k} .$

For $n = 2$ this pair is the first two square formulas again. The product definition of $(k α)$ did not need $α$ to be a whole number. The finite sum (0.17) still does. The series that drops that requirement is below; it does not stop at $k = n$ .

The entry is the sum of the two above it

The coefficients for $n = 0$ through $n = 5$ are a triangle. Each entry is the sum of the two entries sitting above it:

$n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 11111112345136101410151$

The board takes one entry of the $n = 5$ row and builds it from the $n = 4$ row. $K$ stays on $1$ through $4$ , where both shoulders exist. The left bar is the entry above and to the left, the middle bar is the entry above and to the right, and the right bar is those two heights stacked, in the same two colors. At the start $K$ is at $2$ , so the shoulders are $4$ and $6$ and the stack is $10$ . An end of the row, where one shoulder is missing, is not a place $K$ can sit.

Drag K. It stays on 1 through 4. The left bar is the n=4 entry above-left, the middle bar is above-right, and the right bar stacks those two heights into the n=5 entry.

The addition that the bars are drawing is (0.19), and it is not limited to the positive integers on the triangle. For a real or complex $α$ ,

$(k α) + (k + 1 α) = (k + 1 α + 1) .$

(0.19) is the rule the triangle is built from. The arrangement is named for Blaise Pascal (1623–1662), who built an adding machine at twenty; the programming language Pascal is named for him. The same rows for $n = 1$ through $8$ are already in Chu Shih-Chieh's 1303 treatise Siyuan yujian. The triangle is also called Jia Xian's.

Two further sums use the same coefficients. The first runs the upper index in step with the lower:

$(0 α) + (1 α + 1) + \dots + (k α + k) = (k α + k + 1) .$

The second multiplies coefficients from two upper indices and adds along a diagonal:

$(0 α) (k β) + (1 α) (k - 1 β) + \dots + (k α) (0 β) = (k α + β) .$

For a natural number $n$ and an index $k$ with $0 \leq k \leq n$ , the product definition collapses to a factorial quotient, and the row is symmetric:

$(k n) = (n - k n) = \frac{n !}{k ! ( n - k )!} .$

Here $n! = 1 \cdot 2 \dots n$ . Move $K$ from $2$ to $3$ and the two shoulder heights trade places while the stack stays $10$ . That is this symmetry for the middle of the $n = 5$ row. The factorial form needs $k$ between $0$ and $n$ inclusive. The product that opened this page does not.

Set $α = β = k = n$ in the diagonal sum. Symmetry rewrites each product $(j n) (n - j n)$ as a square, and

$(0 n)^{2} + (1 n)^{2} + \dots + (n n)^{2} = (n 2 n) .$

The binomial theorem itself, at two special pairs, counts the whole row. For $a = b = 1$ ,

$(0 n) + (1 n) + \dots + (n n) = 2^{n},$

because the left side is $(1 + 1)^{n}$ . For $a = - b = 1$ , so $a = 1$ and $b = - 1$ , the same theorem gives $(1 + (- 1))^{n}$ . For $n = 1, 2, \dots$ that power is $0$ , and

$(0 n) - (1 n) + (2 n) - \dots + (- 1)^{n} (n n) = 0.$

The $n = 5$ row in the table adds to $32$ , and the alternating signs cancel. Drop the hypothesis that $n$ is a positive integer and neither evaluation is a finite row anymore.

The same triangle writes the first few powers without a new idea. The shares for $n = 3, 4, 5$ are

$(a + b)^{3} (a + b)^{4} (a + b)^{5} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}, = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}, = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5} .$

The exponent leaves the positive integers

Isaac Newton (1643–1727), at twenty-four, found by intuitive reasoning that the same product is the coefficient of an infinite series. For a real exponent $α$ ,

$(1 + x)^{α} = 1 + (1 α) x + (2 α) x^{2} + (3 α) x^{3} + \dots = \sum_{k = 0}^{\infty} (k α) x^{k} .$

That is (0.18). If $α$ is one of $1, 2, \dots$ , then $(k α) = 0$ once $k > α$ , the series stops, and (0.18) is the binomial theorem again.

The board freezes $α = 1/2$ and looks at the left end of $∣ x ∣ < 1$ , where four terms still miss the curve. The sample point stays inside that disk. Four terms of the product definition are

$(1 1/2) = \frac{1}{2}, (2 1/2) = - \frac{1}{8}, (3 1/2) = \frac{1}{16}, (4 1/2) = - \frac{5}{128},$

so the polynomial under $Q$ is $1 + x /2 - x^{2} /8 + x^{3} /16 - 5 x^{4} /128$ .

Drag t. α is frozen at 1/2, and this view stays near the left end of |x|<1. P is (1+x)^α. Q is the sum of (0.18) through the fourth power. The segment is the tail those four terms leave out. Drag toward the right and the gap shrinks.

The segment from $P$ to $Q$ is whatever those four terms omit. At the start it is wide enough to see; drag $t$ to the right and it shortens. That shortening is not a proof that the omitted tail goes to zero. Leonhard Euler, in 1774, at sixty-seven and more than a hundred years after Newton found the series, proved the convergence: for every real $α$ and every complex $x$ with $∣ x ∣ < 1$ , the series (0.18) converges. Outside that disk the board does not go, and the page does not claim the sum.

More than two summands

(0.17) has two summands. The same counting, with a factor for each summand, extends the power. For three,

$(a + b + c)^{2} (a + b + c)^{3} = a^{2} + b^{2} + c^{2} + 2 ab + 2 a c + 2 b c, = a^{3} + b^{3} + c^{3} + 3 a^{2} b + 3 a^{2} c + 3 b^{2} c + 6 ab c + 3 a b^{2} + 3 a c^{2} + 3 b c^{2} .$

For nonzero real or complex numbers $a_{1}, \dots, a_{N}$ and a natural number $n = 1, 2, \dots$ ,

$(a_{1} + a_{2} + \dots + a_{N})^{n} = \sum_{m_{1} + \dots + m_{N} = n} \frac{n !}{m _{1} ! m _{2} ! \dots m _{N} !} a_{1}^{m_{1}} a_{2}^{m_{2}} \dots a_{N}^{m_{N}} .$

The sum runs over every $N$ -tuple of natural numbers from $0$ through $n$ whose sum is $n$ . For $N = 2$ the factorial quotient is $(m _{2} n)$ , and the sum is (0.17) again. What drops, if one $a_{i}$ is zero, is the hypothesis under which this general line is stated.

The object on this page is the share of each product in a power of a sum. For two summands and a positive integer $n$ , that share is $(k n)$ , built by adding the two entries above it, and the square shows the case $n = 2$ only while the sides stay positive. For a real exponent the same product continues as a series, and $∣ x ∣ < 1$ is the condition under which the tail on the last board is claimed to vanish. The question that board can still answer is how the gap from $P$ to $Q$ changes as $t$ moves inside that disk.