Further Maths Proof by induction Practice Questions

Free Further Maths Proof by induction practice questions with full step-by-step worked solutions. Covers series-evaluate, induction, series, summation. Practise exam-style problems and check your method.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
It can be proved by induction that ∑r=1n[r]=n(n+1)2\sum_{r=1}^{n}\left[r\right]=\frac{n (n + 1)}{2} for all positive integers nn. Use this result to evaluate ∑r=110[r]\sum_{r=1}^{10}\left[r\right].
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Worked solution

  1. Quote the proved closed form

    ∑r=1n[r]=n(n+1)2\sum_{r=1}^{n}\left[r\right]=\frac{n (n + 1)}{2}

    The induction has already established this formula.

  2. Substitute n=10n=10 into the closed form

    n=10:55n=10:\quad 55

    Replace nn by the required upper limit.

  3. State the proposition to be proved

    P(n): ∑r=1n[r]=n(n+1)2P(n):\ \sum_{r=1}^{n}\left[r\right]=\frac{n (n + 1)}{2}

    Name the statement so the base case and inductive step can refer to it.

  4. State the value of the sum

    ∑r=110[r]=55\sum_{r=1}^{10}\left[r\right]=55

    This is the required value of the series.

Answer
5555
Question 2
2 markseasy
The matrix M=(2003)M=\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}. Which of the following is M3M^{3}?
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Worked solution

  1. Recall the induction result for MnM^{n}

    Mn=(2n003n)M^{n}=\begin{pmatrix} 2^{n} & 0 \\ 0 & 3^{n} \end{pmatrix}

    Induction establishes a formula for every power.

  2. Substitute n=3n=3

    M3=(80027)M^{3}=\begin{pmatrix} 8 & 0 \\ 0 & 27 \end{pmatrix}

    Evaluate each entry at the required power.

  3. State the proposition to be proved

    P(n): Mn=(2n003n)P(n):\ M^{n}=\begin{pmatrix} 2^{n} & 0 \\ 0 & 3^{n} \end{pmatrix}

    Name the statement about the nnth power of the matrix.

  4. Select the matching matrix

    (80027)\begin{pmatrix} 8 & 0 \\ 0 & 27 \end{pmatrix}

    This is M3M^{3}.

Answer
(80027)\begin{pmatrix} 8 & 0 \\ 0 & 27 \end{pmatrix}
Question 3
4 marksintermediate
In a proof by induction that ∑r=1n[r3]=n2(n+1)24\sum_{r=1}^{n}\left[r^{3}\right]=\frac{n^{2} (n + 1)^{2}}{4}, which of the following is the term uk+1u_{k+1} added when moving from n=kn=k to n=k+1n=k+1?
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Worked solution

  1. Identify the general term

    ur=r3u_r=r^{3}

    The summand defines the rrth term.

  2. Substitute r=k+1r=k+1 and expand

    uk+1=(k+1)3=k3+3k2+3k+1u_{k+1}=(k + 1)^{3}=k^{3} + 3 k^{2} + 3 k + 1

    The added term is the summand evaluated at r=k+1r=k+1.

  3. State the proposition to be proved

    P(n): ∑r=1n[r3]=n2(n+1)24P(n):\ \sum_{r=1}^{n}\left[r^{3}\right]=\frac{n^{2} (n + 1)^{2}}{4}

    Name the statement so the base case and inductive step can refer to it.

  4. Evaluate the left-hand side for the base case

    n=1:LHS=1n=1:\quad \text{LHS}=1

    Substitute n=1n=1 into the summation.

  5. Evaluate the right-hand side for the base case

    n=1:RHS=1n=1:\quad \text{RHS}=1

    Substitute n=1n=1 into the closed form.

  6. Select the added term

    k3+3k2+3k+1k^{3} + 3 k^{2} + 3 k + 1

    This is the term that appears in the inductive step.

Answer
k3+3k2+3k+1k^{3} + 3 k^{2} + 3 k + 1
Question 4
6 markshard
The matrix M=(5−82−3)M=\begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix}. Which of the following is M4M^{4}?
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Worked solution

  1. Recall the induction result for MnM^{n}

    Mn=(4n+1−8n2n1−4n)M^{n}=\begin{pmatrix} 4 n + 1 & - 8 n \\ 2 n & 1 - 4 n \end{pmatrix}

    Induction establishes a formula for every power.

  2. Substitute n=4n=4

    M4=(17−328−15)M^{4}=\begin{pmatrix} 17 & -32 \\ 8 & -15 \end{pmatrix}

    Evaluate each entry at the required power.

  3. State the proposition to be proved

    P(n): Mn=(4n+1−8n2n1−4n)P(n):\ M^{n}=\begin{pmatrix} 4 n + 1 & - 8 n \\ 2 n & 1 - 4 n \end{pmatrix}

    Name the statement about the nnth power of the matrix.

  4. Test the base case n=1n=1

    M1=(5−82−3)M^{1}=\begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix}

    The first power is just the matrix itself.

  5. Substitute n=1n=1 into the proposed formula

    (4n+1−8n2n1−4n)∣n=1=(5−82−3)\begin{pmatrix} 4 n + 1 & - 8 n \\ 2 n & 1 - 4 n \end{pmatrix}\Big|_{n=1}=\begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix}

    The formula reproduces MM, so the base case holds.

  6. State the inductive hypothesis

    Assume Mk=(4k+1−8k2k1−4k)\text{Assume } M^{k}=\begin{pmatrix} 4 k + 1 & - 8 k \\ 2 k & 1 - 4 k \end{pmatrix}

    Assume the formula is correct for some positive integer kk.

  7. Use the power law for matrices

    Mk+1=MkMM^{k+1}=M^{k}M

    One extra factor of MM takes the kkth power to the (k+1)(k+1)th.

  8. Substitute the inductive hypothesis

    Mk+1=(4k+1−8k2k1−4k)(5−82−3)M^{k+1}=\begin{pmatrix} 4 k + 1 & - 8 k \\ 2 k & 1 - 4 k \end{pmatrix}\begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix}

    Replace MkM^{k} by the assumed formula.

  9. Multiply the two matrices

    Mk+1=(4k+5−8k−82k+2−4k−3)M^{k+1}=\begin{pmatrix} 4 k + 5 & - 8 k - 8 \\ 2 k + 2 & - 4 k - 3 \end{pmatrix}

    Multiply row by column and simplify each entry.

  10. Write the target formula at n=k+1n=k+1

    Target=(4k+5−8k−82k+2−4k−3)\text{Target}=\begin{pmatrix} 4 k + 5 & - 8 k - 8 \\ 2 k + 2 & - 4 k - 3 \end{pmatrix}

    Replace nn by k+1k+1 in the proposed formula.

  11. Select the matching matrix

    (17−328−15)\begin{pmatrix} 17 & -32 \\ 8 & -15 \end{pmatrix}

    This is M4M^{4}.

Answer
(17−328−15)\begin{pmatrix} 17 & -32 \\ 8 & -15 \end{pmatrix}
Question 5
8 markschallenging
The matrix M=(3103)M=\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}. Which of the following is M5M^{5}?
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Worked solution

  1. Recall the induction result for MnM^{n}

    Mn=(3n3n−1n03n)M^{n}=\begin{pmatrix} 3^{n} & 3^{n - 1} n \\ 0 & 3^{n} \end{pmatrix}

    Induction establishes a formula for every power.

  2. Substitute n=5n=5

    M5=(2434050243)M^{5}=\begin{pmatrix} 243 & 405 \\ 0 & 243 \end{pmatrix}

    Evaluate each entry at the required power.

  3. State the proposition to be proved

    P(n): Mn=(3n3n−1n03n)P(n):\ M^{n}=\begin{pmatrix} 3^{n} & 3^{n - 1} n \\ 0 & 3^{n} \end{pmatrix}

    Name the statement about the nnth power of the matrix.

  4. Test the base case n=1n=1

    M1=(3103)M^{1}=\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}

    The first power is just the matrix itself.

  5. Substitute n=1n=1 into the proposed formula

    (3n3n−1n03n)∣n=1=(3103)\begin{pmatrix} 3^{n} & 3^{n - 1} n \\ 0 & 3^{n} \end{pmatrix}\Big|_{n=1}=\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}

    The formula reproduces MM, so the base case holds.

  6. State the inductive hypothesis

    Assume Mk=(3k3k−1k03k)\text{Assume } M^{k}=\begin{pmatrix} 3^{k} & 3^{k - 1} k \\ 0 & 3^{k} \end{pmatrix}

    Assume the formula is correct for some positive integer kk.

  7. Use the power law for matrices

    Mk+1=MkMM^{k+1}=M^{k}M

    One extra factor of MM takes the kkth power to the (k+1)(k+1)th.

  8. Substitute the inductive hypothesis

    Mk+1=(3k3k−1k03k)(3103)M^{k+1}=\begin{pmatrix} 3^{k} & 3^{k - 1} k \\ 0 & 3^{k} \end{pmatrix}\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}

    Replace MkM^{k} by the assumed formula.

  9. Multiply the two matrices

    Mk+1=(3⋅3k3kk+3k03⋅3k)M^{k+1}=\begin{pmatrix} 3 \cdot 3^{k} & 3^{k} k + 3^{k} \\ 0 & 3 \cdot 3^{k} \end{pmatrix}

    Multiply row by column and simplify each entry.

  10. Write the target formula at n=k+1n=k+1

    Target=(3⋅3k3kk+3k03⋅3k)\text{Target}=\begin{pmatrix} 3 \cdot 3^{k} & 3^{k} k + 3^{k} \\ 0 & 3 \cdot 3^{k} \end{pmatrix}

    Replace nn by k+1k+1 in the proposed formula.

  11. Compare the product with the target

    (3⋅3k3kk+3k03⋅3k)=(3⋅3k3kk+3k03⋅3k)\begin{pmatrix} 3 \cdot 3^{k} & 3^{k} k + 3^{k} \\ 0 & 3 \cdot 3^{k} \end{pmatrix}=\begin{pmatrix} 3 \cdot 3^{k} & 3^{k} k + 3^{k} \\ 0 & 3 \cdot 3^{k} \end{pmatrix}

    Every entry agrees, so the inductive step is complete.

  12. Deduce the inductive step

    P(k)⇒P(k+1)P(k)\Rightarrow P(k+1)

    Truth at n=kn=k forces truth at n=k+1n=k+1.

  13. Combine the base case and the inductive step

    P(1) true and P(k)⇒P(k+1)P(1)\ \text{true and}\ P(k)\Rightarrow P(k+1)

    Both requirements of induction are satisfied.

  14. State the conclusion of the induction

    Mn=(3n3n−1n03n)∀n∈Z+M^{n}=\begin{pmatrix} 3^{n} & 3^{n - 1} n \\ 0 & 3^{n} \end{pmatrix}\quad \forall n\in\mathbb{Z}^{+}

    The formula holds for every positive integer power.

  15. Recall the structure of a proof by induction

    P(1) true,P(k)⇒P(k+1)P(1)\ \text{true},\quad P(k)\Rightarrow P(k+1)

    A proof by induction needs a base case and an inductive step.

  16. Select the matching matrix

    (2434050243)\begin{pmatrix} 243 & 405 \\ 0 & 243 \end{pmatrix}

    This is M5M^{5}.

Answer
(2434050243)\begin{pmatrix} 243 & 405 \\ 0 & 243 \end{pmatrix}

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