\[\mathrm{f}(x) = (1 + 2x)^6 (1 - x)^4\] Find the first three terms, in ascending powers of \(x\), of the binomial expansion of \((1 + 2x)^6\). (3) Find the...

Eleven marks on a product of two binomial expansions, ending with a numerical estimate. The strategy is to expand each bracket separately as far as \(x^2\), then multiply the two series keeping only the terms up to \(x^2\). Part (d) then exploits the fact that a truncated series is an excellent approximation when \(x\) is small.

(a) The first three terms of \((1 + 2x)^6\) [3 marks]

\[(1 + 2x)^6 = 1 + \binom{6}{1}(2x) + \binom{6}{2}(2x)^2 + \ldots \quad \textbf{B1}\] \[= 1 + 6(2x) = 1 + 12x \quad \textbf{B1}\] \[+ \ 15(4x^2) = +\ 60x^2 \quad \textbf{B1}\]

So \((1 + 2x)^6 = 1 + 12x + 60x^2 + \ldots\). Three separate B marks, one for the structure and one for each of the \(x\) and \(x^2\) terms, so a slip in one does not cost the others. The point to watch is \((2x)^2 = 4x^2\): the 2 is squared as well, which is why the coefficient is \(15 \times 4 = 60\) and not 15.

(b) The first three terms of \((1 - x)^4\) [2 marks]

\[(1 - x)^4 = 1 + \binom{4}{1}(-x) + \binom{4}{2}(-x)^2 + \ldots = 1 - 4x \quad \textbf{B1}\] \[+ \ 6x^2 \quad \textbf{B1}\]

The alternating signs come from the powers of \(-x\): odd powers are negative and even powers positive, so \((-x)^2 = +x^2\) and the \(x^2\) coefficient is \(+6\), not \(-6\).

(c) Show the expansion of \(\mathrm{f}(x)\) begins \(1 + 8x + 18x^2\) [3 marks]

The result is printed, so the marks are for the multiplication. Collect the products whose indices add to each required power: \[\mathrm{f}(x) = \bigl(1 + 12x + 60x^2 + \ldots\bigr)\bigl(1 - 4x + 6x^2 + \ldots\bigr)\] \[\text{Constant: } 1 \times 1 = 1 \quad \textbf{B1}\] \[\text{Term in } x: \ 1(-4) + 12(1) = -4 + 12 = 8 \quad \textbf{B1}\] \[\text{Term in } x^2: \ 1(6) + 12(-4) + 60(1) = 6 - 48 + 60 = 18 \quad \textbf{B1}\] There are exactly three ways to produce an \(x^2\) term, and all three must appear: constant times \(x^2\), \(x\) times \(x\), and \(x^2\) times constant. Terms in \(x^3\) and beyond are ignored because only the first three are wanted, so there is no need to expand either bracket further. Note the near-cancellation in the \(x^2\) coefficient: \(6 - 48 + 60\) is only 18, so dropping any one contribution changes the answer dramatically.

(d) An estimate for \((1.02)^6 \times (0.99)^4\) [3 marks]

Match the numerical bases to the algebraic ones. With \(x = 0.01\), \[1 + 2x = 1 + 0.02 = 1.02 \quad \text{and} \quad 1 - x = 1 - 0.01 = 0.99 \quad \textbf{B1}\] so \(\mathrm{f}(0.01) = (1.02)^6 \times (0.99)^4\) exactly. Substituting into the truncated expansion: \[\text{Estimate} = 1 + 8(0.01) + 18(0.01)^2 \quad \textbf{M1}\] \[= 1 + 0.08 + 0.0018 = 1.0818 \ \text{(4 d.p.)} \quad \textbf{A1}\] The B1 is for identifying that \(x = 0.01\) reproduces both bases, which is the insight the part is testing. Note \((0.01)^2 = 0.0001\), so the \(x^2\) term contributes \(0.0018\); the successive terms shrink by roughly a factor of 100, which is exactly why three terms suffice for four decimal places.

Common wrong turns on this question

• Writing \((2x)^2\) as \(2x^2\) in part (a), giving \(30x^2\) instead of \(60x^2\).
• Making the \(x^2\) coefficient in part (b) negative. Squaring \(-x\) gives a positive result.
• Finding only two of the three \(x^2\) contributions in part (c), most often omitting \(12 \times (-4)\).
• Choosing \(x = 0.02\) in part (d) because \(1.02\) is visible. The algebraic form is \(1 + 2x\), so \(2x = 0.02\) and \(x = 0.01\); the same \(x\) must also satisfy \(1 - x = 0.99\), which confirms the choice.
• Writing \((0.01)^2\) as \(0.001\) and reporting \(1.098\).
• Rounding to fewer than four decimal places.

How to check

Evaluate the target directly: \((1.02)^6 = 1.126162\ldots\) and \((0.99)^4 = 0.960596\ldots\), whose product is \(1.081787\ldots\), which rounds to \(1.0818\). The three-term estimate therefore agrees to all four decimal places, confirming both the expansion and the substitution. A structural check on part (c) is also worth doing: the constant term of \(\mathrm{f}(x)\) must be \(1^6 \times 1^4 = 1\), and the \(x\) coefficient must be \(6(2) + 4(-1) = 8\), matching the answer.