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Grade 11-12 (age 16-18)

high school Binomial Theorem

Expand (a + b)ⁿ and every term, aⁿ, aⁿ⁻¹b and so on, carries a coefficient; it counts the ways to take b from r of the n brackets, which is the combination C(n, r). Lined up row by row, these coefficients form Pascal's triangle, where every number is the sum of the two just above it. To get one particular coefficient you don't have to expand everything: in the general term C(n, r) aⁿ⁻ʳbʳ, find the single r that gives the right exponents. Move the exponent slider n and row n of the triangle lights up, while the same numbers appear as the coefficients of the expansion below it.

The Coefficients Come from Combinations

4
👀 Triangle and expansion
①(a + b)ⁿ is (a + b) multiplied n times, so it contains every product made by taking a or b from each bracket
②There are C(n, r) ways to take b from r of the brackets, so the coefficient of aⁿ⁻ʳbʳ is C(n, r)
③Writing these coefficients row by row from n = 0 gives Pascal's triangle
④As the two lines in the figure show, each number is the sum of the two above it, which is the identity C(n, r) = C(n − 1, r − 1) + C(n − 1, r)

The Binomial Theorem

Binomial theorem
(a+b)n = ∑r=0n C(n,r) an-r br
C(n, r) counts the ways to take b from r of the n brackets
General term
Tr+1 = C(n,r) an-r br
The (r + 1)th term for r = 0, 1, …, n, used to compute one particular coefficient directly

Finding a Particular Coefficient

🎯 Match the exponents first
①Choose r so that the term has the exponent you want
②Put that r into the general term and compute the coefficient
③For the constant term, find the r that makes the exponent of the variable 0
④When a or b is an expression with a number in it, such as 2x or −1, the powers of that number go into the coefficient too

Worked Examples

Example 1
Find the coefficient of x² in the expansion of (x + 2)⁴.
1
The general term is C(4, r) x⁴⁻ʳ 2ʳ. For x² we need 4 − r = 2, so r = 2.
2
Put r = 2 in and compute the coefficient.
C(4,2)·22 = 6·4 = 24
▸ 24
Set the exponent of the variable equal to the degree you want and fix r first; the rest is arithmetic.
Example 2
Find the coefficient of x³ in the expansion of (2x − 1)⁵.
1
The general term is C(5, r) (2x)⁵⁻ʳ (−1)ʳ. For x³ we need 5 − r = 3, so r = 2.
2
With r = 2, the cube of 2 and the square of −1 are multiplied in as well.
C(5,2)·23·(−1)2 = 10·8·1 = 80
▸ 80
Leave out the powers of 2 and −1 inside the brackets and you get the wrong answer, 10.

Summary

Key results
(a+b)n = ∑ C(n,r) an-r br, general term C(n,r) an-r br
The coefficients are the combinations C(n, r); for one term, match r in the general term
2021 CSAT Math type, adapted
Find the constant term in the expansion of (x + 1/x)⁶.
①6
②15
③20
④30
⑤45
▸ ③ 20
1
The general term is C(6, r) x⁶⁻ʳ (1/x)ʳ = C(6, r) x⁶⁻²ʳ.
2
The constant term has x to the power 0, so 6 − 2r = 0 and r = 3.
C(6,3) = 20
🎯 Exam Points
①The coefficients of (a + b)ⁿ are the combinations C(n, r)
②For a particular term, match the exponents in the general term C(n, r) aⁿ⁻ʳbʳ to fix r
③The constant term comes from the r that makes the exponent of the variable 0
④Include the powers and signs of constants inside the brackets in the coefficient
⑤Pascal's triangle is a quick table of coefficients for small n
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