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DSE M2 Mathematical Induction Practice

DSE M2 · 60 practice questions with instant marking and worked solutions.

Mathematical Induction is an important part of the DSE M2 mathematics curriculum. This topic develops the base case, the inductive hypothesis and the inductive step for sums, divisibility and inequalities. Begin by identifying what every number, symbol, diagram and unit represents. Translate the information into a mathematical relationship before calculating. Each line of working should have a purpose. Finally use estimation, substitution, an inverse operation, or a second representation to check the result. This habit separates a misunderstanding from a copying error and makes correction more useful.

A reliable approach to Mathematical Induction has four stages. First, mark the given information and the quantity required. Second, choose an appropriate representation, property or formula. Third, carry out the operations in a clear order while retaining signs and units. Fourth, return to the original context and decide whether the result is possible. For a graph or table, read the title, scale and labels before using a value. For algebra, inspect brackets, negative signs and the order of operations before simplifying.

The 60 original questions in this topic use a shared mathematical specification in Chinese and English. Difficulty rises from level 1 to level 5. Early questions establish the language and direct skills of Mathematical Induction; middle questions add two-step reasoning, conversions and common variations; later questions connect ideas and require a deliberate check. In multiple-choice work, calculate before looking at the choices. In fill-in and short-answer work, retain decisive steps. For true-or-false statements, support the decision with a calculation or counterexample. 12 of them are DSE Extended Part style structured questions answered in parts (a), (b), (c), where later parts build on earlier ones.

Effective revision is more than recording a score. After completing the Mathematical Induction set, group errors by concept tags and decide whether each problem arose from interpretation, forming an expression, calculation, or presenting the answer. Retry the item on a later day without opening the solution, then test the same method with a different value. If the same tag appears repeatedly, return to the explanation and rebuild the concept. If signs or units are the issue, use a fixed pre-submission checklist.

Mathematical Induction is closely linked with the rest of the Extended Part: algebraic and trigonometric skills underpin calculus, and calculus in turn supports probability distributions and statistical inference. A learner who can explain aloud why every step is valid has moved beyond imitation. Estimate a reasonable range, complete an exact solution, and compare the two. Agreement between context, estimate and exact answer is strong evidence that the solution is sound.

This batch adds 20 questions, bringing the topic to 60. The new questions cover controlled variations of different concepts, with paired bilingual conditions, hints, worked solutions and reproducible checks.

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  1. 1 Basic
    It can be proved by induction that $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$ for all positive integers $n$. Find the value of $1^2+2^2+\dots+6^2$. Choose the correct answer.
  2. 2 Basic
    Mathematical induction is used to prove that $5^n-1$ is divisible by 4 for every positive integer $n$. Find the value of the expression in the base case $n=1$. Give the answer only.

  3. 3 Basic
    Induction is used to prove $\sum_{r=1}^{n}r\cdot 2^{r}=(n-1)2^{n+1}+2$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=6$. Show the essential working and give the answer.

  4. 4 Basic
    The inequality $n!>2^n$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. The least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$) is 3. Is this true?
  5. 5 Basic
    It can be proved by induction that $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$ for all positive integers $n$. Find the value of $1^3+2^3+\dots+13^3$. Choose the correct answer.
  6. 6 Basic
    Mathematical induction is used to prove that $7^n+5$ is divisible by 6 for every positive integer $n$. Find the value of the expression in the base case $n=1$. Give the answer only.

  7. 7 Basic
    Induction is used to prove $\sum_{r=1}^{n}r\cdot 2^{r}=(n-1)2^{n+1}+2$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=9$. Show the essential working and give the answer.

  8. 8 Basic
    The inequality $n!>3^n$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. The least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$) is 6. Is this true?
  9. 9 Basic
    It can be proved by induction that $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$ for all positive integers $n$. Find the value of $1^3+2^3+\dots+6^3$. Choose the correct answer.
  10. 10 Basic
    Mathematical induction is used to prove that $7^n-1$ is divisible by 6 for every positive integer $n$. Find the value of the expression in the base case $n=1$. Give the answer only.

  11. 11 Basic
    Induction is used to prove $\sum_{r=1}^{n}r\cdot 2^{r}=(n-1)2^{n+1}+2$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=8$. Show the essential working and give the answer.

  12. 12 Basic
    The inequality $3^n>n^3$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. The least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$) is 3. Is this true?
  13. 13 Standard
    It can be proved by induction that $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$ for all positive integers $n$. Find the value of $1^3+2^3+\dots+8^3$. Choose the correct answer.
  14. 14 Standard
    Mathematical induction is used to prove that $9^n-1$ is divisible by 8 for every positive integer $n$. In the inductive step, $9^{k+1}-1$ is written as $9(9^k-1)+q$. Find the value of the constant $q$. Give the answer only.

  15. 15 Standard
    Induction is used to prove $\sum_{r=1}^{n}r\cdot 2^{r}=(n-1)2^{n+1}+2$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=5$. Show the essential working and give the answer.

  16. 16 Standard DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1+3+5+\dots+(2n-1)=n^2$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=14$ to $n=15$.
    3. (c) Find $S(14)$.
  17. 17 Standard
    It can be proved by induction that $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$ for all positive integers $n$. Find the value of $1^2+2^2+\dots+7^2$. Choose the correct answer.
  18. 18 Standard
    Mathematical induction is used to prove that $5^n+3$ is divisible by 4 for every positive integer $n$. In the inductive step, $5^{k+1}+3$ is written as $5(5^k+3)+q$. Find the value of the constant $q$. Give the answer only.

  19. 19 Standard
    Induction is used to prove $\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=11$. Show the essential working and give the answer.

  20. 20 Standard DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1\times 2+2\times 3+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=12$ to $n=13$.
    3. (c) Find $S(12)$.
  21. 21 Standard
    It can be proved by induction that $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$ for all positive integers $n$. Find the value of $1^2+2^2+\dots+20^2$. Choose the correct answer.
  22. 22 Standard
    Mathematical induction is used to prove that $5^n-1$ is divisible by 4 for every positive integer $n$. In the inductive step, $5^{k+1}-1$ is written as $5(5^k-1)+q$. Find the value of the constant $q$. Give the answer only.

  23. 23 Standard
    Induction is used to prove $\sum_{r=1}^{n}r(r+1)=\frac{n(n+1)(n+2)}{3}$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=8$. Show the essential working and give the answer.

  24. 24 Standard DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=11$ to $n=12$.
    3. (c) Find $S(11)$.
  25. 25 Advanced
    It can be proved by induction that $\sum_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}$ for all positive integers $n$. Find the value of $\frac{1}{1\times 2}+\dots+\frac{1}{19\times20}$ (as a fraction in lowest terms). Choose the correct answer.
  26. 26 Advanced
    It can be proved by induction that $1+3+5+\dots+(2n-1)=n^2$ for all positive integers $n$. Find the value of $1+3+\dots+35$. Give the answer only.

  27. 27 Advanced
    Mathematical induction is used to prove that $6^n+4$ is divisible by 5 for every positive integer $n$. In the inductive step, $6^{k+1}+4$ is written as $6(6^k+4)+q$. The value of the constant $q$ is -20. Is this true?
  28. 28 Advanced DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=12$ to $n=13$.
    3. (c) Find $S(12)$.
  29. 29 Advanced
    The inequality $2^n>n^2$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. Find the least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$). Choose the correct answer.
  30. 30 Advanced
    It can be proved by induction that $1+3+5+\dots+(2n-1)=n^2$ for all positive integers $n$. Find the value of $1+3+\dots+39$. Give the answer only.

  31. 31 Advanced
    Mathematical induction is used to prove that $9^n+7$ is divisible by 8 for every positive integer $n$. In the inductive step, $9^{k+1}+7$ is written as $9(9^k+7)+q$. The value of the constant $q$ is -56. Is this true?
  32. 32 Advanced DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=10$ to $n=11$.
    3. (c) Find $S(10)$.
  33. 33 Advanced
    The inequality $n!>3^n$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. Find the least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$). Choose the correct answer.
  34. 34 Advanced
    It can be proved by induction that $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$ for all positive integers $n$. Find the value of $1^2+2^2+\dots+10^2$. Give the answer only.

  35. 35 Advanced
    Mathematical induction is used to prove that $4^n+2$ is divisible by 3 for every positive integer $n$. The value of the expression in the base case $n=1$ is 6. Is this true?
  36. 36 Advanced DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=8$ to $n=9$.
    3. (c) Find $S(8)$.
  37. 37 Challenging
    Induction is used to prove $\sum_{r=1}^{n}r\cdot 2^{r}=(n-1)2^{n+1}+2$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=9$. Choose the correct answer.
  38. 38 Challenging
    The inequality $2^n>2n+1$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. Find the least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$). Show the essential working and give the answer.

  39. 39 Challenging
    It can be proved by induction that $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$ for all positive integers $n$. The value of $1^3+2^3+\dots+18^3$ is 29259. Is this true?
  40. 40 Challenging DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=14$ to $n=15$.
    3. (c) Find $S(14)$.
  41. 41 Challenging
    Induction is used to prove $\sum_{r=1}^{n}\frac{r(r+1)}{2}=\frac{n(n+1)(n+2)}{6}$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=12$. Choose the correct answer.
  42. 42 Challenging
    It can be proved by induction that $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$ for all positive integers $n$. Find the value of $1^3+2^3+\dots+8^3$. Show the essential working and give the answer.

  43. 43 Challenging
    It can be proved by induction that $1+3+5+\dots+(2n-1)=n^2$ for all positive integers $n$. The value of $1+3+\dots+9$ is 30. Is this true?
  44. 44 Challenging DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1\times 2+2\times 3+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=7$ to $n=8$.
    3. (c) Find $S(7)$.
  45. 45 Challenging
    Induction is used to prove $\sum_{r=1}^{n}\frac{r(r+1)}{2}=\frac{n(n+1)(n+2)}{6}$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=10$. Choose the correct answer.
  46. 46 Challenging
    The inequality $n!>2^n$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. Find the least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$). Show the essential working and give the answer.

  47. 47 Challenging
    It can be proved by induction that $\sum_{k=1}^{n}k^3=\left[\frac{n(n+1)}{2}\right]^2$ for all positive integers $n$. The value of $1^3+2^3+\dots+10^3$ is 3035. Is this true?
  48. 48 Challenging DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=13$ to $n=14$.
    3. (c) Find $S(13)$.
  49. 49 Elite
    Mathematical induction is used to prove that $3^n+1$ is divisible by 2 for every positive integer $n$. In the inductive step, $3^{k+1}+1$ is written as $3(3^k+1)+q$. Find the value of the constant $q$. Give the answer only.

  50. 50 Elite
    Induction is used to prove $\sum_{r=1}^{n}r\cdot 2^{r}=(n-1)2^{n+1}+2$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=3$. Show the essential working and give the answer.

  51. 51 Elite
    The inequality $2^n>n^2$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. The least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$) is 5. Is this true?
  52. 52 Elite DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1\times 2+2\times 3+\dots+n(n+1)=\frac{n(n+1)(n+2)}{3}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=9$ to $n=10$.
    3. (c) Find $S(9)$.
  53. 53 Elite
    Mathematical induction is used to prove that $4^n+2$ is divisible by 3 for every positive integer $n$. In the inductive step, $4^{k+1}+2$ is written as $4(4^k+2)+q$. Find the value of the constant $q$. Give the answer only.

  54. 54 Elite
    Induction is used to prove $\sum_{r=1}^{n}(2r-1)^2=\frac{n(2n-1)(2n+1)}{3}$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=9$. Show the essential working and give the answer.

  55. 55 Elite
    The inequality $n!>2^n$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. The least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$) is 4. Is this true?
  56. 56 Elite DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1^2+2^2+\dots+n^2=\frac{n(n+1)(2n+1)}{6}$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=9$ to $n=10$.
    3. (c) Find $S(9)$.
  57. 57 Elite
    Mathematical induction is used to prove that $9^n-1$ is divisible by 8 for every positive integer $n$. Find the value of the expression in the base case $n=1$. Give the answer only.

  58. 58 Elite
    Induction is used to prove $\sum_{r=1}^{n}\frac{r(r+1)}{2}=\frac{n(n+1)(n+2)}{6}$. Assuming the result for $n=k$, the $(k+1)$th term is added to both sides to reach $n=k+1$. Find the value of the term added when $k=11$. Show the essential working and give the answer.

  59. 59 Elite
    The inequality $2^n>2n+1$ holds for all integers $n\ge N$, which can be proved by induction with $n=N$ as the base case. The least suitable positive integer $N$ (the inequality holds at $n=N$ but not at $n=N-1$) is 3. Is this true?
  60. 60 Elite DSE long question
    It can be proved by induction that, for all positive integers $n$, $S(n)=1+3+5+\dots+(2n-1)=n^2$. Answer each part; give the answer only.
    1. (a) Base case: find the value of the right-hand side when $n=1$.
    2. (b) Inductive step: find the term added to both sides when going from $n=11$ to $n=12$.
    3. (c) Find $S(11)$.

The report lists what you got wrong and why. Only you can see it.

Common mistakes

Common mistakes in Mathematical Induction

This topic needs particular care with the base case, the inductive hypothesis and the inductive step for sums, divisibility and inequalities. The first mistake is applying a remembered rule before reading all conditions. The second is compressing working so much that a sign, bracket, decimal point or unit disappears. The third is accepting an impossible answer without checking: for example, a probability above one, a negative physical length, an angle outside the geometric constraints, or a result far from the estimate. Separate the conditions, model, calculation and check, and finish by stating what the answer represents.

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