Binomial expansion for rational n and series modelling – Year 13 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Mark scheme: 1 mark for taking out the factor \(2^{-3}\), 1 mark for the correct coefficients \(-3\) and \(6\), 1 mark for simplifying, 1 mark for the range.

   \((2+x)^{-3}=2^{-3}\left(1+\dfrac x2\right)^{-3}=\dfrac18\left(1-3\cdot\dfrac x2+\dfrac{(-3)(-4)}2\cdot\dfrac{x^2}4\right)=\dfrac18\left(1-\dfrac{3x}2+\dfrac{3x^2}2\right)\). So the expansion is \(\dfrac18-\dfrac{3x}{16}+\dfrac{3x^2}{16}\) (3 marks). Valid for \(\left|\dfrac x2\right|\lt 1\), i.e. \(|x|\lt 2\) (1 mark).

   Answer: \(\dfrac1{(2+x)^3}=\dfrac18-\dfrac{3x}{16}+\dfrac{3x^2}{16}+\cdots\) for \(-2\lt x\lt 2\).

   Check with \(x=0.1\): \(\dfrac1{2.1^3}=0.10798\), while the expansion gives \(0.125-0.01875+0.001875=0.108125\), close as expected for a small \(x\).

2. Answer to Question 2 (4 marks)

   Mark scheme: 2 marks for the expansion (1 for the method with \(n=\tfrac12\), 1 for all terms correct), 1 mark for the range, 1 mark for the estimate.

   1. Use \(n=\tfrac12\) and inside term \(u=-4x\), so \(u^2=16x^2\) and \(u^3=-64x^3\). \((1+u)^{1/2}=1+\dfrac u2-\dfrac{u^2}8+\dfrac{u^3}{16}\) with \(u=-4x\): \(1-2x-2x^2-4x^3+\cdots\). Every term after the first is negative.
   2. The inside term is \(-4x\). \(|4x|\lt 1\), so \(|x|\lt \tfrac14\). So the expansion is valid for \(-\tfrac14\lt x\lt \tfrac14\).
   3. Since \(1-4(0.01)=0.96\), substitute \(x=0.01\). \(x=0.01\): \(1-0.02-0.0002-0.000004=0.979796\). Calculator: 0.979796. So \(\sqrt{0.96}\approx0.979796\).

3. Answer to Question 3 (4 marks)

   Mark scheme: 2 marks for the partial fractions (1 per constant), 1 mark for the series, 1 mark for the range.

   1. Multiply both sides by \((1-x)(2+x)\) and substitute convenient values. \(4-x=A(2+x)+B(1-x)\). \(x=1\): \(3=3A\), \(A=1\). \(x=-2\): \(6=3B\), \(B=2\). So \(h(x)=\dfrac1{1-x}+\dfrac2{2+x}\). Check at \(x=0\): \(1+1=2\) and \(\tfrac{4}{2}=2\).
   2. Expand each fraction with \(n=-1\), taking out the factor 2 from \(2+x\). \(\dfrac1{1-x}=1+x+x^2+x^3\) and \(\dfrac2{2+x}=\left(1+\dfrac x2\right)^{-1}=1-\dfrac x2+\dfrac{x^2}4-\dfrac{x^3}8\). Adding: \(2+\dfrac x2+\dfrac54x^2+\dfrac78x^3+\cdots\). Constant term: \(1+1=2\); \(x\) term: \(1-\tfrac12=\tfrac12\).
   3. The whole series needs both conditions at the same time. The first needs \(|x|\lt 1\) and the second \(|x|\lt 2\), so valid for \(|x|\lt 1\). The narrower condition wins.

4. Answer to Question 4 (4 marks)

   Mark scheme: 1 mark for the two equations, 1 mark for \(a\) and \(n\), 1 mark for the coefficient, 1 mark for the range.

   1. Compare the general coefficients \(na\) and \(\tfrac{n(n-1)}2a^2\) with the given values, then divide. \(na=-3\) and \(\dfrac{n(n-1)a^2}2=\dfrac{27}4\), so \((n-1)a=\dfrac{27/2}{-3}=-\dfrac92\). Then \(a=na-(n-1)a=-3+\dfrac92=\dfrac32\) and \(n=-2\). Check: \(na=(-2)\times\tfrac32=-3\).
   2. Use \(\tfrac{n(n-1)(n-2)}{3!}a^3\). \(\dfrac{(-2)(-3)(-4)}{6}\left(\dfrac32\right)^3=-4\times\dfrac{27}8=-\dfrac{27}2\). Valid for \(\left|\dfrac32x\right|\lt 1\), so \(|x|\lt \tfrac23\). So the coefficient of \(x^3\) is \(-\tfrac{27}2\).

5. Answer to Question 5 (4 marks)

   Mark scheme: 1 mark for the expansion, 1 mark for substituting \(x=0.04\), 1 mark for the estimate, 1 mark for the percentage.

   1. Use \(n=-\tfrac12\) with inside term \(-x\): the coefficients are \(\tfrac12\) and \(\tfrac38\). \((1-x)^{-1/2}=1+\dfrac x2+\dfrac{3x^2}8\). With \(x=0.04\): \(1+0.02+0.0006=1.0206\). Calculator: \(0.96^{-1/2}=1.020621\); the estimate is 1.02060 (to 5 d.p.). The next term, \(\tfrac{5}{16}x^3\), would only add about 0.00002.
   2. The period is multiplied by about 1.0206, so subtract 1 and convert to a percentage. The period rises by about 2.1%. So reducing gravity by 4% makes the pendulum about 2.1% slower.

Total : 20 marks