The Disputed Binomial Coefficient
Two students expand (2 + 3x)⁶. Student A writes the general term as C(6, r)·2^(6−r)·(3x)^r and claims the coefficient of x³ is C(6,3)·2³·3³ = 20·8·27 = 4320. Student B instead writes the general term as C(6, r)·2^r·3^(6−r)·x^(6−r) and claims the coefficient of x³ is C(6,3)·2³·3³ = 4320 as well, so both agree. A third student points out that the coefficient of x⁴ should be checked too, and that the expansion's coefficients should follow a predictable ratio pattern. The teacher asks the group to justify which general term is correct and to verify using the ratio of consecutive coefficients.
1. Determine which general term is correct and compute the coefficients of x³ and x⁴ in (2 + 3x)⁶.
Read the worked answer
The correct general term is C(6, r)·2^(6−r)·(3x)^r, because the constant 2 is raised to the power of the number of times it is chosen (6 − r) while 3x is chosen r times. So the coefficient of x^r is C(6, r)·2^(6−r)·3^r.
- Coefficient of x³ (r = 3): C(6,3)·2³·3³ = 20·8·27 = 4320.
- Coefficient of x⁴ (r = 4): C(6,4)·2²·3⁴ = 15·4·81 = 4860.
Student B's form happens to give the same number for x³ only because C(6,3) is symmetric and the powers swap, but it labels the wrong power of x, so it is not generally correct.
2. Use the ratio of consecutive coefficients to confirm the values found and identify the largest coefficient in the expansion.
Read the worked answer
For the coefficient a_r = C(6,r)·2^(6−r)·3^r, the ratio a_(r+1)/a_r = [(6−r)/(r+1)]·(3/2).
Check a₄/a₃ = [(6−3)/(3+1)]·(3/2) = (3/4)(3/2) = 9/8. Indeed 4320 × 9/8 = 4860. ✓
Now find where the ratio crosses 1. Set [(6−r)/(r+1)]·(3/2) = 1 → 3(6−r) = 2(r+1) → 18 − 3r = 2r + 2 → 16 = 5r → r = 3.2. So coefficients increase up to r = 3 and decrease after r = 4, meaning a₄ is the largest. Computing a₅ = C(6,5)·2·3⁵ = 6·2·243 = 2916 < 4860 confirms a₄ = 4860 is the maximum coefficient.