Multiplying out (x + 2)⁴ by hand takes three rounds of FOIL. The triangle does it in one line.
Find the coefficient
Find the coefficient of x³ in (2x − 3)⁵
The x³ term needs (2x)³(−3)², so it is C(5,2) × 2³ × (−3)² = 10 × 8 × 9 = 720. It is positive because the −3 is squared. D keeps a minus that the even power has already killed; A leaves out the coefficient C(5,2) = 10.
Each entry is the sum of the two above it. Row n gives the expansion of power n.
| Power | Coefficients | Terms |
|---|---|---|
| 2 | 1, 2, 1 | 3 |
| 3 | 1, 3, 3, 1 | 4 |
| 4 | 1, 4, 6, 4, 1 | 5 |
| 5 | 1, 5, 10, 10, 5, 1 | 6 |
A power of n always produces n + 1 terms, and the coefficients read the same backwards as forwards.
(x + 2)⁴
coefficients 1, 4, 6, 4, 1
x⁴, x³, x², x, 1
the first part, counting down
2⁰, 2¹, 2², 2³, 2⁴
the second, counting up
1, 2, 4, 8, 16
work those out first
1·x⁴ + 4·x³·2 + 6·x²·4 + 4·x·8 + 1·16
multiply each pair
x⁴ + 8x³ + 24x² + 32x + 16
tidied
The third term is 6 · x² · 2², which is 24x², not 6x² and not 12x². In every term the exponents of the two parts must add to 4 — that is the fastest self-check there is.
The same pattern evaluates a power of a decimal exactly. 1.2³ = (1 + 0.2)³ = 1 + 3(0.2) + 3(0.2)² + (0.2)³ = 1 + 0.6 + 0.12 + 0.008 = 1.728.
Nothing new is happening: the second part is (−2), and a negative raised to an odd power stays negative. Carrying the minus inside the bracket handles it automatically.
term = C(n, r) · an−r · br
r counts from 0, so the (r + 1)th term.
(2x − 3)⁵, want x³
a = 2x, b = −3, n = 5
an−r = (2x)³ → n − r = 3, so r = 2
the exponent picks r
C(5, 2)(2x)³(−3)²
into the formula
10 · 8x³ · 9
C(5,2) = 10, (2x)³ = 8x³
= 720x³, coefficient 720
positive — (−3) is squared
Answering 10 gives only the Pascal number, and ignoring the 2 in (2x)³ loses a factor of 8. The coefficient is the whole numerical part: 10 × 8 × 9.
The first term has r = 0, so r is always one less than the term number.
(x + y)⁷, 4th term → r = 3
4 − 1
C(7, 3) x⁴ y³
7 − 3 = 4 on the x
C(7, 3) = 7 · 6 · 53 · 2 · 1 = 35
three factors top and bottom
= 35x⁴y³
exponents add to 7 ✓
When x sits in both parts, write the general term, collect the power of x, and choose r so that power is 0. Find the constant term of (x² − 1x)⁶.
C(6, r)(x²)6−r(−1x)r
a = x², b = −1x
= C(6, r)(−1)r x12 − 2r · x−r
1xr is x−r
power of x: 12 − 3r = 0 → r = 4
no x left
C(6, 4)(−1)4 = 15 × 1 = 15
it is the 5th term
Solving the power equation must give a whole r from 0 to n. If it gives r = 2.5, the expansion has no constant term — that is a legitimate answer.
(2x + 1)⁴ has 5 terms → middle is the 3rd, r = 2
n even: middle is term n2 + 1
C(4, 2)(2x)²(1)² = 6 × 4x² = 24x²
the middle term
sum of coefficients of (3x − 1)⁴ = (3 − 1)⁴ = 16
put x = 1 — no expanding
x² coefficient of (1 + x)n is C(n, 2) = n(n − 1)2 = 45
given a coefficient, find n
n(n − 1) = 90 = 10 × 9 → n = 10
two consecutive whole numbers
When n is odd there are two middle terms, numbers n + 12 and n + 32. For the coefficient sum, x = 1 works because every power of 1 is 1, leaving only the coefficients.
Binomial items almost never ask for a full expansion. They ask for one coefficient, the kth or middle term, the constant term, or n. Write the general term C(n, r)an−rbr straight away, pick r from the power asked for, and remember the signs and numbers inside the brackets are raised too.
Then the expansion must equal the bracket, worked out directly.
Find the coefficient
Find the coefficient of x² in (3x + 1)⁵
x² comes from (3x)²(1)³, giving C(5,3) × 3² = 10 × 9 = 90. A forgets to square the 3; B squares the wrong thing and uses 3³. The power on the bracket tells you which term, the power on 3x tells you the number.
