Further Mathematics

1. 1.

   Use partial fractions to find sum from r = 1 to 10 of 1/[r(r + 1)].

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Write the first two and last two differences to see the cancellation pattern.
   2. The last negative denominator is 11, although the final index is 10.

   Marking points

   • 1/[r(r + 1)] = 1/r - 1/(r + 1).
   • All interior terms cancel, leaving 1 - 1/11.
   • The sum is 10/11.

   Examiner tip: Keep the last negative term when telescoping.

2. 2.

   A geometric series has terms 12, -6, 3, ... . Find its sum to infinity and the exact error after four terms.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Alternation does not stop convergence because the ratio magnitude is below one.
   2. Add the first four terms 12 - 6 + 3 - 3/2, then subtract that sum from 8.

   Marking points

   • The ratio is -1/2, whose modulus is less than 1.
   • The infinite sum is 8.
   • S4 = 15/2; the signed error S_infinity - S4 is 1/2.

   Examiner tip: Use the signed ratio in 1 - r, not its absolute value.

3. 3.

   Prove by induction that sum from r = 1 to n of r^2 equals n(n + 1)(2n + 1)/6 for n >= 1.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Put the added square over denominator 6 and factor out k + 1.
   2. The remaining quadratic 2k^2 + 7k + 6 factors as (k + 2)(2k + 3).

   Marking points

   • At n = 1 both sides equal 1.
   • Assume the formula at k and add (k + 1)^2.
   • Factor to (k + 1)(k + 2)(2k + 3)/6, the formula at k + 1.
   • The base and implication establish the result for all positive integers.

   Examiner tip: State the induction assumption rather than using the target formula for k + 1.

4. 4.

   Prove that 7^n - 1 is divisible by 6 for every integer n >= 1 using induction.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Rewrite the next power to expose the exact expression in the hypothesis.
   2. The extra 6 is essential: 7(7^k - 1) alone is six less than the required expression.

   Marking points

   • n = 1 gives 6, divisible by 6.
   • Assume 7^k - 1 = 6m for an integer m.
   • 7^(k + 1) - 1 = 7(7^k - 1) + 6 = 6(7m + 1).
   • Since 7m + 1 is an integer, the implication and base prove the claim.

   Examiner tip: Checking many integers is evidence, not a proof for every n.

5. 5.

   u1 = 1 and u(n + 1) = 2u(n) + 3. Conjecture a formula for u(n), prove it by induction, then find sum u(r) for r = 1,...,n.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Shift by three: v(n) = u(n) + 3 satisfies v(n + 1) = 2v(n) with v1 = 4.
   2. Sum the geometric v terms and subtract three for each of the n terms.

   Marking points

   • First terms 1, 5, 13 suggest u(n) = 2^(n + 1) - 3.
   • Base gives 1; substitution gives 2[2^(k + 1) - 3] + 3 = 2^(k + 2) - 3, completing induction.
   • The sum is 4(2^n - 1) - 3n.

   Examiner tip: The constant subtraction contributes -3n to the sum, not just -3.

6. 6.

   Prove 2^n > n^2 for every integer n >= 5 by induction. Explain why starting at n = 4 cannot establish the strict inequality.

   [4 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. To bound the quadratic difference, write k^2 - 2k - 1 = (k - 1)^2 - 2 >= 14 for k >= 5.
   2. Use this bound after doubling the hypothesis; a valid induction step still requires a valid base.

   Marking points

   • Base: 32 > 25 at n = 5.
   • Assuming 2^k > k^2 gives 2^(k + 1) > 2k^2.
   • 2k^2 - (k + 1)^2 = k^2 - 2k - 1 > 0 for k >= 5, completing induction.
   • At n = 4 both sides are 16, so the strict base fails.

   Examiner tip: Preserve > throughout; an equality at the base cannot prove a strict claim.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.