Binomial expansion and sequences – Year 12 A level maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   Use the binomial expansion with \(a=2\), \(b=-3x\) and \(n=4\); the coefficients are 1, 4, 6.

   \(16+4\times8(-3x)+6\times4(9x^2)=16-96x+216x^2\) (3). \(16-0.96+0.0216=15.0616\) (1).

   Since \(2-3x=1.97\) we need \(x=0.01\). The estimate is \(1.97^4\approx15.0616\); the calculator gives \(15.0614\ldots\), so the estimate is good. Marks: 1 for each correct term, 1 for the substitution.

   Common errors: forgetting to raise 2 to the powers 3 and 2, or losing the minus sign of \(-3x\), which turns the \(x\) coefficient into \(+96\). The \(x^2\) coefficient is positive because \((-3x)^2=9x^2\).

2. Answer to Question 2 (4 marks)

   Expand \((1+2x)^5\) as far as \(x^3\); the \(x^3\) term of the product comes from two pairs: \(80x^3\times1\) and \(40x^2\times(-x)\).

   \((1+2x)^5=1+10x+40x^2+80x^3+\dots\) (1). Multiply by \((1-x)\): the \(x^3\) term is \(80x^3-40x^3=40x^3\) (2 for the products, 1 for the answer 40).

   So the coefficient of \(x^3\) is 40. Marks: 1 for the expansion, 2 for the two correct products, 1 for the final answer.

3. Answer to Question 3 (4 marks)

   1. Write \(u_3=a+2d=11\) and \(u_8=a+7d=31\), then subtract to eliminate \(a\). \(5d=20\), \(d=4\), \(a=3\) (2) Check: \(3+2\times4=11\) ✓ and \(3+7\times4=31\) ✓. The first term is 3 and the common difference is 4.
   2. The 20th term is \(3+19\times4=79\), and \(S_{20}=\tfrac{20}2[2\times3+19\times4]\). \(S_{20}=10(6+76)=820\) (2) So the sum of the first 20 terms is 820.

   Alternative for (b): \(S_n=\tfrac n2(a+l)\) with last term \(l=79\) gives \(10\times(3+79)=820\), the same total. The method mark is awarded for a correct sum formula with \(a=3\) and \(d=4\) substituted.

4. Answer to Question 4 (4 marks)

   1. Since \(|r|=\tfrac23\lt 1\) the series converges and \(S_\infty=\dfrac a{1-r}\). \(\dfrac{18}{1/3}=54\) (2) The sum to infinity is 54.
   2. Use \(S_n=54\left(1-\left(\tfrac23\right)^n\right)\) and take logarithms; dividing by \(\ln\tfrac23\lt 0\) reverses the inequality. \(54\left(1-\left(\tfrac23\right)^n\right)\gt 53\) gives \(\left(\tfrac23\right)^n\lt \tfrac1{54}\) (1), \(n\gt 9.84\) so \(n=10\) (1) Because \(n\) must be a whole number, the smallest value is \(n=10\). Check: \(\left(\tfrac23\right)^{10}\approx0.0173\lt \tfrac1{54}\approx0.0185\), but \(\left(\tfrac23\right)^9\approx0.0260\) is too large ✓.

   In (a), the first mark is for quoting the formula with \(a=18\) and \(r=\tfrac23\), the second for 54. In (b), trial and improvement is also accepted if both \(S_9\approx52.60\) and \(S_{10}\approx53.06\) are shown. Since 53 is very close to the limit 54, many terms are needed before the partial sums pass it.

5. Answer to Question 5 (4 marks)

   1. The weekly amounts form an arithmetic sequence with \(a=2\) and \(d=1\), so \(u_{25}=2+24\times1\). £26 (1) In week 25 the child saves £26.
   2. Use \(S_n=\tfrac n2(a+l)\). \(\tfrac{25}2(2+26)=£350\) (1) After 25 weeks the total is £350.
   3. Form \(S_n=\tfrac n2[4+(n-1)]=\tfrac n2(n+3)\) and test whole numbers near the solution of \(n^2+3n=1000\). \(S_n=\tfrac n2(n+3)\gt 500\), so \(n^2+3n\gt 1000\) (1); \(n=30\) gives 990, \(n=31\) gives 1054, so 31 weeks (1) The total first exceeds £500 after 31 weeks.

Total : 20 marks