You can find the eighth term of an expansion without writing the other twelve.
Find the coefficient
In the expansion of (x + 2)⁵, what is the coefficient of x³?
The term with x³ is C(5,2) x³ 2² = 10 × x³ × 4 = 40x³. A is the binomial coefficient alone and forgets that the 2 is raised to a power too. The exponents must add to 5, so x³ pairs with 2².
Each entry is the sum of the two above it. Row 5 reads 1, 5, 10, 10, 5, 1 — and those are exactly the coefficients of (a + b)⁵.
To go further, start and end the next row with 1 and add each neighbouring pair of the row above: 1 + 5, 5 + 10, 10 + 10, and so on. Row n always has n + 1 entries, so (a + b)ⁿ has n + 1 terms: (a + b)⁵ has six, and (a + b)⁹ has ten.
row 3: 1, 3, 3, 1
the coefficients
x³, x²(2), x(2²), 2³
x goes down, 2 goes up
x³ + 3(2)x² + 3(4)x + 8
multiply through
= x³ + 6x² + 12x + 8
the expansion
Writing x³ + 3x²(2) + 3x(2) + 2 loses every power on the 2 after the first. Each term's two exponents must add to n, so check that on every line: 3 + 0, 2 + 1, 1 + 2, 0 + 3.
Writing out twenty rows of Pascal's triangle to expand (x + 1)²⁰ is possible but foolish. The combination formula reaches any row directly.
term r + 1 = C(n, r) an−r bʳ
note the r + 1
3rd term means r = 2
one less than the position
C(7,2) x⁵ 2² = 21 × x⁵ × 4
substitute
= 84x⁵
the 3rd term
The off-by-one is the whole difficulty: the FIRST term has r = 0. Subtract one from the position before anything else and the rest is substitution.
The same general term, C(n, r) an − r bʳ, with a or b less simple than before.
A minus sign: the coefficient of x² in (x − 3)⁵
C(5, r) x5 − r (−3)ʳ
b is −3, sign included
5 − r = 2, so r = 3
match the power
C(5,3)(−3)³ = 10 × (−27) = −270
an odd power keeps the minus
Keep the sign inside the bracket. Odd r gives a negative term, even r a positive one.
A coefficient on x: the coefficient of x⁴ in (2x + 1)⁶
C(6, r) (2x)6 − r 1ʳ
a is 2x, not x
6 − r = 4, so r = 2
match the power
C(6,2) × 2⁴ = 15 × 16 = 240
the 2 is raised too
Raising 2x to the 4th gives 16x⁴. Answering 15 forgets the 2 altogether.
A constant term: (x + 1x)⁶
C(6, r) x6 − r (1x)ʳ = C(6, r) x6 − 2r
the powers of x combine
constant means 6 − 2r = 0, so r = 3
x⁰ = 1
C(6,3) = 20
the 4th term
"Constant term" means the power of x is 0. Combine the powers first, then set them to 0.
general term: C(9, r) x9−r 1ʳ
write it out
want 9 − r = 5
so r = 4
C(9,4) = 126
same as C(9,5)
the term is 126x⁵
the 5th term
Setting the exponent equal to the power you want and solving for r turns "find the term containing x⁵" into a one-line equation. It is the same technique whatever the binomial, however large n is.
Tests ask for one coefficient or one term — "the coefficient of x³", "the 4th term", "the constant term" — rarely the full expansion. The fastest route: write the general term C(n, r) an − r bʳ, set the power of x to what is wanted, solve for r, and substitute, raising both parts. A 10-second check: the two exponents must add to n, and the term's position is r + 1.
Find the coefficient
In the expansion of (x + 3)⁴, what is the coefficient of x²?
The exponents add to 4, so x² pairs with 3². The term is C(4,2) x² 3² = 6 × x² × 9 = 54x². A is the binomial coefficient alone, and C uses 3 rather than 3² — forgetting that the second part is raised to a power as well.
Hint: if x has exponent 2 and the total must be 4, the 3 carries the other 2. Row 4 of Pascal's triangle is 1, 4, 6, 4, 1.
