Pascal's Triangle Row 12: All Entries and Row Sum

The binomial theorem, expanded term by term

(2x - 3)⁴ comes out as 16x⁴ - 96x³ + 216x² - 216x + 81, and this binomial expansion calculator shows every one of those five terms with the coefficient, the two powers and the product that built it. Type the two terms as you would write them, 2x, -3, x^2 or 1/x, pick the power n, and optionally name a power of x to get its coefficient on its own. The same tool gives the binomial coefficient C(n, k) exactly up to n = 1000 and draws Pascal's triangle up to row 30. Nothing is rounded: fractions stay fractions, and a 30-digit coefficient keeps all 30 digits.

n + 1

terms in (a + b)ⁿ before like terms merge

2ⁿ

sum of the coefficients in row n

C(n, k) = C(n, n - k)

every row reads the same both ways

Typing a binomial the calculator can read

What do you want to calculate - an expansion, a single binomial coefficient C(n, k), or a block of Pascal's triangle.
First term and second term - one number and at most one letter each, with an optional power: 2x, -3, x^2, 3/2y, 0.5x. For a letter below the line write 1/x or 2/x^3. A minus sign belongs to the term, so (x - 2)⁵ is x and -2.
Power n - a whole number from 0 to 60 for an expansion, up to 1000 for C(n, k), up to 30 for the triangle.
Power of x to look up - optional. Type 4 to get the coefficient of x⁴, or 0 for the constant term. It works when the binomial has one letter.
k - for C(n, k) only, from 0 to n.
Read the result: the expanded polynomial, the tiles (your coefficient, the sum of all coefficients, the constant term, the middle term) and the table with one row per term.

What the theorem says, and why its numbers are counts

Multiply (a + b) by itself n times and every term of the result comes from choosing, in each of the n brackets, either a or b. A term with k b's and n - k a's can be picked in as many ways as there are to choose which k brackets supply the b. That count is the binomial coefficient C(n, k), read "n choose k", so (a + b)ⁿ = Σ C(n, k) a^n-k b^k for k from 0 to n.

The general term, written T_k+1 = C(n, k) a^n-k b^k, is what exam questions usually target. Finding the coefficient of x⁴ in (3x + 2)⁷ means finding the k where the power of x is 4, here k = 3, and multiplying C(7, 3) = 35 by 3⁴ and 2³. The answer is 22,680, and the calculator returns it in the lookup tile without writing out the other seven terms by hand.

When a term carries a negative power, as in (x + 1/x)⁶, the powers of x run 6, 4, 2, 0, -2, -4, -6. The term with x⁰ is the constant term, a favorite of textbook exercises, and here it is C(6, 3) = 20.

Pascal's triangle, rows 0 to 10

Each entry is the sum of the two above it. Row n holds the coefficients of (a + b)ⁿ, and the last column shows that each row adds up to twice the one before.

Row n	Coefficients of (a + b)ⁿ	Row sum, 2ⁿ
0	1	1
1	1, 1	2
2	1, 2, 1	4
3	1, 3, 3, 1	8
4	1, 4, 6, 4, 1	16
5	1, 5, 10, 10, 5, 1	32
6	1, 6, 15, 20, 15, 6, 1	64
7	1, 7, 21, 35, 35, 21, 7, 1	128
8	1, 8, 28, 56, 70, 56, 28, 8, 1	256
9	1, 9, 36, 84, 126, 126, 84, 36, 9, 1	512
10	1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1	1,024

Expansions and coefficients that come up in exercises

Each row was produced by the calculator and checked against the polynomial multiplied out bracket by bracket.

Binomial	What is asked	Answer	Interpretation
(x + 2)⁵	full expansion	x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32	row 5 times powers of 2
(2x - 3)⁴	full expansion	16x⁴ - 96x³ + 216x² - 216x + 81	signs alternate because the second term is negative
(3x + 2)⁷	coefficient of x⁴	22,680	35 × 81 × 8
(x + 1/x)⁶	constant term	20	the middle term, x³ · x^-3
(x² + 1/x)⁹	coefficient of x³, constant term	126 and 84	the powers step by 3: 18, 15, ..., 0, -3, ...
(a + b)³	two letters	a³ + 3a²b + 3ab² + b³	no constant term
(x + 1)⁴ at x = 10	the value	14,641 = 11⁴	row 4 read as digits, 1 4 6 4 1

Patterns hidden in the coefficients

Symmetry. C(n, k) = C(n, n - k), because choosing k items to take is the same as choosing n - k to leave. That is why every row of the triangle reads the same from either end, and why the calculator shows C(n, n - k) next to your C(n, k).

Row sums are powers of 2. Setting a = b = 1 in (a + b)ⁿ gives 2ⁿ, so row 10 adds up to 1,024. The same trick, letting every letter equal 1, gives the "sum of all coefficients" tile: for (2x - 3)⁴ it is (2 - 3)⁴ = 1.

Powers of 11. Rows 0 to 4 read as numbers are 1, 11, 121, 1331 and 14,641, the powers of 11. From row 5 on the entries reach two digits and carry into each other, so the pattern needs the carries done by hand.

Odd entries and diagonals. Row n has 2 to the power of (the number of 1s in n written in binary) odd entries: row 5 is 101 in binary, so 4 odd entries (1, 5, 5, 1). And the shallow diagonals of the triangle add up to the Fibonacci numbers 1, 1, 2, 3, 5, 8.

How to read the result

The headline is the whole polynomial, with fractions in brackets before a letter so that (3/2)x cannot be misread as 3/(2x). The table keeps the terms in the order k = 0, 1, 2, ..., which is the order of the general term T_k+1, so the fourth row is T₄. If both terms use the same letter, terms with the same power are added and the headline says so. In C(n, k) mode the number is exact at any size: C(10, 3) = 120, the count of 5-card poker hands C(52, 5) = 2,598,960, and C(60, 30) = 118,264,581,564,861,424, eighteen digits that a spreadsheet would round in its last places. The coin-flip tile translates C(n, k) into the chance of exactly k heads in n fair flips, C(n, k) / 2ⁿ.

Binomial expansion questions, short answers

How do you find a specific term without expanding everything?

Write the general term C(n, k) a^n-k b^k, set its power of x equal to the one you want and solve for k. For x⁴ in (3x + 2)⁷, k = 3 and the coefficient is 22,680. The lookup box does this for you.

How do I enter (x - 2)^5 or (1 - 3x)^4?

Put the minus sign on the term: first term x, second term -2; or first term 1, second term -3x. The order of the two terms only changes the order of the rows, not the polynomial.

What is the constant term, and what if there is none?

It is the term with no letter left, x⁰. In (x + 1/x)⁶ it is 20. In (x + 1/x)⁵ the powers are all odd, so there is none; the lookup for power 0 then returns 0 and lists the powers that do occur.

Does it work for fractional or negative powers such as (1 + x)^-1?

No. With a negative or fractional n the binomial series never ends, and the calculator expands only whole powers from 0 to 60. Negative powers inside a term, such as 1/x or x^-2, are fine.

Why is the middle coefficient the largest?

Going along a row, each entry is the previous one times (n - k) / (k + 1), which is above 1 until k reaches about n / 2. So row 10 climbs to 252 at k = 5 and falls back. With other numbers in the terms, as in (2x - 3)⁴, the largest coefficient can move away from the middle.

Is "n choose k" the same as the combinations formula?

Yes, C(n, k) = n! / (k! (n - k)!) counts unordered selections, which is why the coefficients of an expansion are also counts of choices. For permutations, repetition and ordered selections, use the combinations and permutations calculator.

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Powers & Roots Calculator

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What are the numbers in row 12 of Pascal's triangle?

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