Math Fair Poster on a Binomial Expansion
For the school math fair, a student has built a display around the expansion of (2x² - 1/x)⁹. Her poster makes three claims: (i) the expansion contains a constant term; (ii) the coefficient of x² is zero, 'because no term can produce x²'; and (iii) the sum of all the coefficients in the expansion is 2⁹. Visitors are also invited to find the term containing x⁶ without writing out all ten terms. A judge who has to sign off on the poster asks for a written verification of each claim, with the general term shown explicitly, and for the actual value of the constant term.
1. Write the general term of the expansion and use it to verify claim (i), giving the value of the constant term.
Read the worked answer
With a = 2x², b = -1/x and n = 9, the general term is:
T(k+1) = C(9,k)(2x²)^(9-k)(-1/x)^k = C(9,k) 2^(9-k) (-1)^k x^(2(9-k)) x^(-k) = C(9,k) 2^(9-k) (-1)^k x^(18-3k)
A constant term requires the exponent 18 - 3k to be 0, so k = 6, which is an integer between 0 and 9. The claim is true and the term exists:
T₇ = C(9,6) 2³ (-1)⁶ = 84 × 8 × 1 = 672
The constant term is 672. Note that all ten powers in the expansion are 18, 15, 12, 9, 6, 3, 0, -3, -6, -9, so the term count (10) matches the index count.
2. Evaluate claim (ii) about the coefficient of x², and as a bonus find the term in x⁶ the visitors are asked for.
Read the worked answer
Claim (ii) reaches a true conclusion with an invalid reason. Require 18 - 3k = 2, which gives k = 16/3. Since k must be a whole number from 0 to 9, no such term exists, and the coefficient of x² really is 0. But the justification 'no term can produce x²' is not an argument at all; the acceptable reason is the exponent-arithmetic one, showing that only multiples of 3 offset by 18 can appear. The judge should require the corrected reasoning even though the coefficient is right.
For x⁶: 18 - 3k = 6 gives k = 4, so T₅ = C(9,4) 2⁵ (-1)⁴ x⁶ = 126 × 32 × 1 x⁶ = 4032x⁶
The judgement being tested is that a correct number obtained from a faulty method is not full credit on a 'verify' task.
3. Evaluate claim (iii) about the sum of the coefficients, and explain the reasoning that justifies the method.
Read the worked answer
Claim (iii) is false. The sum of the coefficients of a polynomial is its value at x = 1, because substituting x = 1 turns every term into its own coefficient:
(2(1)² - 1/1)⁹ = (2 - 1)⁹ = 1⁹ = 1
So the coefficients sum to 1, not to 2⁹ = 512. The value x = 1 is valid since the expression is defined there (only x = 0 is excluded).
A quick consistency check: the constant term alone is 672, which already exceeds the total 1, so the negative coefficients must more than cancel it. That is reasonable, since terms with odd k carry a negative sign and grow quickly. If the poster wanted a sum of 2⁹ it would have to display (2x² + 1/x) evaluated at a different point; the claim conflates the values at x = 1 and at x = 0 or uses the binomial coefficients alone, ignoring the 2^(9-k) and sign factors.