Example 1: Finding a coefficient
Find the coefficient of x³ in the expansion of (2 − x)⁵.
Tr+1 = C(5, r) 2⁵⁻ʳ (−x)ʳ = C(5, r) 2⁵⁻ʳ (−1)ʳ xʳ- For
x³,r = 3 C(5, 3) × 2² × (−1)³ = 10 × 4 × (−1) = −40
Answer: −40
For a positive integer n, (a + b)ⁿ = Σ C(n, r) aⁿ⁻ʳ bʳ for r = 0 to n. The general term is Tr+1 = C(n, r) aⁿ⁻ʳ bʳ. To find a particular coefficient, write the general term, collect the power of x, and solve for r.
(a + b)ⁿ = C(n, 0)aⁿ + C(n, 1)aⁿ⁻¹b + … + C(n, n)bⁿTr+1 = C(n, r) aⁿ⁻ʳ bʳC(n, r) = n! / (r!(n − r)!)(1 + x)ⁿ = 1 + nx + [n(n − 1)/2]x² + … + xⁿΣ(axr + byr) = aΣxr + bΣyr, Σi=1n a = naFind the coefficient of x³ in the expansion of (2 − x)⁵.
Tr+1 = C(5, r) 2⁵⁻ʳ (−x)ʳ = C(5, r) 2⁵⁻ʳ (−1)ʳ xʳx³, r = 3C(5, 3) × 2² × (−1)³ = 10 × 4 × (−1) = −40Answer: −40
Find the constant term in the expansion of (x² + 1/x)⁶.
Tr+1 = C(6, r) (x²)⁶⁻ʳ (x⁻¹)ʳ = C(6, r) x¹²⁻³ʳ12 − 3r = 0, so r = 4C(6, 4) = 15Answer: 15
(2 − x)⁵ the second term is −x, so the general term contains (−1)ʳ. Dropping it makes every odd-power coefficient the wrong sign.C(n, r) is the (r + 1)th term, because r starts from 0. The "4th term" uses r = 3.How to study: Click any question below to instantly load it on the coordinate grid and check your steps.