Math · College algebra · Worked example

Find one term of a binomial expansion

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

{(2 x + 1)}^{5}

Match the power of x

Each term is C(5, k)(2x)⁵⁻ᵏ(1)ᵏ. The power of x is 5 − k, so x³ needs k = 2.

C(5, 2) = 5!/(2! 3!) = 10, the third entry of row 5: 1, 5, 10, 10, 5, 1.

(\begin{matrix} 5 \\ 2 \end{matrix}) = \frac{5!}{2! 3!} = 10

(2x)³ = 8x³: the 2 is cubed along with x. So the term is 10 · 8x³ · 1 = 80x³.

(\begin{matrix} 5 \\ 2 \end{matrix}) {(2 x)}^{3} {(1)}^{2} = 80 x^{3}

(2x + 1)⁵ = 32x⁵ + 80x⁴ + 80x³ + 40x² + 10x + 1, and its x³ term is 80x³.

The coefficient of x³ is 80.

Find the constant term of (x + 2/x)⁴.

Show the answer and explanation

Each term is C(4, k)x⁴⁻ᵏ(2/x)ᵏ = C(4, k)2ᵏx⁴⁻²ᵏ. The power is 0 when k = 2, giving C(4, 2) · 2² = 6 · 4 = 24.

Open Pascal’s Triangle at row 5, position 2: it shows C(5, 2) = 10 and the matching term 10a³b².

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