Binomial Theorem

When you expand (a+b)ⁿ, where do the coefficients of each term come from? Remarkably, every one of them is a combination C(n, r), exactly matching row n of Pascal's triangle. So the coefficient of the aⁿ⁻ʳbʳ term is captured by the single general term C(n, r), and you can pin down any particular coefficient just by matching the exponents. Move the exponent-n slider to watch row n of Pascal's triangle light up and see that this row is precisely the set of coefficients in the expansion.

Coefficients Come from Combinations

Exponent n4

👀 See It

①Expanding (a+b)ⁿ, each term's coefficient is the combination C(n,r)

②Lining up these coefficients gives Pascal's triangle

③Each number is the sum of the two above it — the same as the combination identity

The Binomial Theorem

Binomial theorem

(a+b)ⁿ = ∑_r=0ⁿ C(n,r) a^n-r b^r

C(n,r) is the combination — the number of ways to pick b r times out of n

General term

T_r+1 = C(n,r) a^n-r b^r

The (r+1)-th term for r=0,1,…,n — used to pinpoint a specific term

Finding the Coefficient of a Term

🎯 Match the Exponent

①Choose r so the exponent of the desired variable matches

②Plug that r into the general term to compute the coefficient

③For the constant term, find the r making the variable exponent 0

Compute It Directly

Example 1

Find the coefficient of x² in the expansion of (x+2)⁴.

The general term is C(4,r) x^4-r 2^r. For x², 4−r=2, so r=2.

Substitute r=2 to compute the coefficient.

C(4,2)·2² = 6·4 = 24

▸ 24

Setting r first via "variable exponent = desired degree" is the key.

Example 2

Find the coefficient of x³ in the expansion of (2x−1)⁵.

The general term is C(5,r) (2x)^5-r (−1)^r. For x³, 5−r=3, so r=2.

Substitute r=2.

C(5,2)·2³·(−1)² = 10·8·1 = 80

▸ 80

Do not drop the powers of the inner constants (2, −1).

Wrap-up

Key result

(a+b)ⁿ = ∑ C(n,r) a^n-r b^r, general term C(n,r) a^n-r b^r

Coefficients are combinations C(n,r) — pinpoint a term by matching 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

The general term is C(6,r) x^6-r (1/x)^r = C(6,r) x^6-2r.

The constant term has exponent 0, so 6−2r=0 gives r=3.

C(6,3) = 20

🎯 Exam Points

①Coefficients of (a+b)ⁿ are combinations C(n,r)

②Use the general term C(n,r)aⁿ⁻ʳbʳ for a specific term

③The constant term has variable exponent 0

④Mind the powers of inner constants

⑤Pascal’s triangle is a quick table of coefficients

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Combination

Meaning of Probability

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