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FP1: ProofEdexcel International A Level Further Maths: Topic test

20 questions, 54 marks

Edexcel International A Level Further Maths

FP1: Proof topic test

Total 54 marks

Name

Class

Date

  1. 1
    Proof by induction is used to show that ∑r=1nr2=n(n+1)(2n+1)6\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6} for all positive integers nn.
    (a)
    The result is assumed true for n=kn=k. Which expression is the sum of the first k+1k+1 square numbers, written using this assumption?
    [1 mark]
    • Ak(k+1)(2k+1)6+k2\frac{k(k+1)(2k+1)}{6}+k^2
    • Bk(k+1)(2k+1)6+(k+1)2\frac{k(k+1)(2k+1)}{6}+(k+1)^2
    • Ck(k+1)(2k+1)6+(k+1)\frac{k(k+1)(2k+1)}{6}+(k+1)
    • D(k+1)(k+2)(2k+3)6+(k+1)2\frac{(k+1)(k+2)(2k+3)}{6}+(k+1)^2
    (b)
    In the inductive step the expression k(k+1)(2k+1)6+(k+1)2\frac{k(k+1)(2k+1)}{6}+(k+1)^2 is written as k+16[  ⋯  ]\frac{k+1}{6}\left[\;\cdots\;\right]. What is the expression in the bracket?
    [1 mark]
    • A2k2+k+62k^2+k+6
    • B2k2+7k+12k^2+7k+1
    • C2k2+13k+122k^2+13k+12
    • D2k2+7k+62k^2+7k+6
    (c)
    Show that the result is true for n=1n=1, and write down the conclusion that completes the proof once the inductive step has been shown.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    Let f(n)=5n+3f(n)=5^n+3 for positive integers nn. It is to be proved by induction that f(n)f(n) is divisible by 44.
    (a)
    Which expression is equal to f(k+1)−f(k)f(k+1)-f(k)?
    [1 mark]
    • A4×5k4\times5^k
    • B5k5^k
    • C4×5k+14\times5^{k+1}
    • D5k+1+35^{k+1}+3
    (b)
    Which statement is the correct inductive hypothesis?
    [1 mark]
    • Af(n)f(n) is divisible by 44 for all positive integers nn.
    • Bf(k+1)f(k+1) is divisible by 44.
    • Cf(k)f(k) is divisible by 44 for some positive integer kk.
    • Df(1)f(1) is divisible by 44.
    (c)
    Show that f(k+1)=5f(k)−12f(k+1)=5f(k)-12.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    A sequence is defined by u1=5u_1=5 and un+1=3un−4u_{n+1}=3u_n-4 for n≥1n\ge1.
    (a)
    Find u2u_2 and u3u_3, and verify that the formula un=3n+2u_n=3^n+2 gives the correct values of u1u_1, u2u_2 and u3u_3.
    [3 marks]
    (b)
    Prove by induction that un=3n+2u_n=3^n+2 for all positive integers nn.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    B=(2101)\mathbf{B}=\begin{pmatrix}2&1\\ 0&1\end{pmatrix}.
    (a)
    Prove by induction that Bn=(2n2n−101)\mathbf{B}^n=\begin{pmatrix}2^n&2^n-1\\ 0&1\end{pmatrix} for all positive integers nn.
    [6 marks]
    (b)
    Use the result in part (a) to answer the following. (i) Write down B10\mathbf{B}^{10}. (ii) Find the least positive integer nn for which the top-right entry of Bn\mathbf{B}^n is greater than 10610^6. (iii) Find det⁡Bn\det\mathbf{B}^n in terms of nn and state what it represents for the transformation Bn\mathbf{B}^n.
    [6 marks]

    Total for question 4: 12 marks

  5. 5
    Let P(n)P(n) be the statement ∑r=1nr(r+1)(r+2)=n(n+1)(n+2)(n+3)4\sum_{r=1}^{n}r(r+1)(r+2)=\frac{n(n+1)(n+2)(n+3)}{4}.
    (a)
    What are the values of the left-hand side and the right-hand side of P(1)P(1)?
    [1 mark]
    • ALHS =6=6 and RHS =24=24
    • BBoth equal 2424
    • CLHS =1=1 and RHS =6=6
    • DBoth equal 66
    (b)
    Assuming P(k)P(k), which expression must be added to k(k+1)(k+2)(k+3)4\frac{k(k+1)(k+2)(k+3)}{4} to give the sum of the first k+1k+1 terms?
    [1 mark]
    • Ak(k+1)(k+2)k(k+1)(k+2)
    • B(k+1)(k+2)(k+3)(k+1)(k+2)(k+3)
    • C(k+2)(k+3)(k+4)(k+2)(k+3)(k+4)
    • D(k+1)(k+2)(k+3)(k+4)4\frac{(k+1)(k+2)(k+3)(k+4)}{4}
    (c)
    Show that k(k+1)(k+2)(k+3)4+(k+1)(k+2)(k+3)=(k+1)(k+2)(k+3)(k+4)4\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)=\frac{(k+1)(k+2)(k+3)(k+4)}{4}.
    [2 marks]

    Total for question 5: 4 marks

  6. 6
    Let h(n)=32n+7h(n)=3^{2n}+7 for positive integers nn. It is to be proved by induction that h(n)h(n) is divisible by 88.
    (a)
    Find h(2)h(2).
    [1 mark]
    • A1616
    • B8181
    • C8888
    • D4343
    (b)
    In the inductive step it is shown that h(k+1)=9h(k)−56h(k+1)=9h(k)-56. Which statement correctly completes the step?
    [1 mark]
    • ASince h(k)h(k) is divisible by 88 by assumption and 56=8×756=8\times7, both terms are multiples of 88, so h(k+1)h(k+1) is divisible by 88.
    • BSince 56=8×756=8\times7 is a multiple of 88, h(k+1)h(k+1) is divisible by 88.
    • CSince h(1)=16h(1)=16 is divisible by 88, h(k+1)h(k+1) is divisible by 88.
    • DSince h(k+1)=9h(k)−56h(k+1)=9h(k)-56 is smaller than 9h(k)9h(k), it is divisible by 88.
    (c)
    Show that h(k+1)=9h(k)−56h(k+1)=9h(k)-56.
    [2 marks]

    Total for question 6: 4 marks

  7. 7
    A sequence is defined by u1=1u_1=1 and un+1=un1+unu_{n+1}=\frac{u_n}{1+u_n} for n≥1n\ge1.
    (a)
    Find u2u_2, u3u_3 and u4u_4, and conjecture a formula for unu_n.
    [3 marks]
    (b)
    Prove by induction that un=1nu_n=\frac1n for all positive integers nn.
    [4 marks]

    Total for question 7: 7 marks

  8. 8
    Let Sn=∑r=1nr×2r=1×2+2×4+3×8+⋯+n×2nS_n=\sum_{r=1}^{n}r\times2^r=1\times2+2\times4+3\times8+\dots+n\times2^n.
    (a)
    Prove by induction that Sn=(n−1)2n+1+2S_n=(n-1)2^{n+1}+2 for all positive integers nn.
    [6 marks]
    (b)
    Use the result in part (a) to find (i) ∑r=610r×2r\sum_{r=6}^{10}r\times2^r and (ii) the least nn for which Sn>106S_n>10^6.
    [6 marks]

    Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).