68 practice questions on Principle of Mathematical Induction , sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the Principle of Mathematical Induction notes .
Induction as a Domino Chain P(1): base case falls P(k) knocks down P(k+1) ...and so on, forever Base case = first domino tipped; inductive step = each domino guaranteed to tip the next one Mathematical induction works like a row of dominoes: proving the base case P(1) tips the first domino, and proving the inductive step (P(k) ⟹ P(k+1)) guarantees each domino knocks over the next - together these two facts guarantee ALL dominoes fall, without checking each one individually.
Easy - 20 questions Q1.
The principle of mathematical induction is used to prove statements for which set of numbers?
A All real numbersB All natural numbersC All irrational numbersD All negative integersShow answer & explanation →
Q2.
What is the first step of a proof by mathematical induction?
A Assume P(k) is true before checking anything elseB Verify the base case, usually P(1)C Prove P(k+1) directly without a base caseD Substitute n equal to infinity into the statementShow answer & explanation →
Q3.
In the inductive step, what do we assume?
A P(n) is already proven true for every natural number nB P(k) is true for some natural number kC P(1) is false, contradicting the base case requirementD P(k+1) is false, which we then aim to disproveShow answer & explanation →
Q5.
If the base case fails for a statement P(n), what can we conclude?
A P(n) remains true for any value of n regardless of the base case outcomeB Induction cannot establish the statement starting from that base caseC P(n) is false for this particular natural number, though that alone proves littleD The inductive step becomes unnecessary here and can safely be skipped overShow answer & explanation →
Q6.
Which best describes the inductive step?
A Proving P(1) only, without considering any later caseB Showing that if P(k) is true, then P(k+1) is also trueC Proving P(n) directly for one specific, large chosen nD Disproving P(k) to show the statement fails in generalShow answer & explanation →
Q7.
Mathematical induction is most commonly compared to which everyday analogy?
A A row of falling dominoesB A game of chessC A balance scaleD A circular race trackShow answer & explanation →
Q8.
To prove n<sup>3</sup> - n is divisible by 6 for all natural numbers n by induction, what is checked first?
A That (k+1)<sup>3</sup> - (k+1) is divisible by 6B That 1<sup>3</sup> - 1 = 0 is divisible by 6C That n is evenD That n<sup>3</sup> is divisible by 6Show answer & explanation →
Q9.
Which closing statement correctly completes an induction proof?
A Hence P(n) is true only for n=1, since that is the only case actually checkedB Hence, by the principle of mathematical induction, P(n) is true for all natural numbers nC Hence P(k) is false, so the inductive step cannot proceed any further from hereD Hence the proof remains incomplete without further verification of every individual caseShow answer & explanation →
Q11.
Mathematical induction is a technique for proving statements about:
A natural numbersB all real numbersC complex numbersD irrational numbersShow answer & explanation →
Q12.
Verifying the statement for the first value (usually n = 1) is called the:
A base caseB inductive stepC final conclusionD mere guessShow answer & explanation →
Q13.
Assuming the statement is true for n = k is called the ___ hypothesis:
A inductiveB the baseC the finalD the randomShow answer & explanation →
Q14.
Proving the statement for n = k + 1 using the assumption is called the:
A inductive stepB base caseC final answerD random guessShow answer & explanation →
Q15.
A successful induction proves the statement for all n greater than or equal to:
A the chosen starting valueB only ten and aboveC one hundred and aboveD positive infinityShow answer & explanation →
Q16.
The principle of mathematical induction is often pictured as a row of falling:
A falling dominoesB loose stonesC stacked bricksD flipped coinsShow answer & explanation →
Q17.
If the base case is not verified, the induction proof is:
A invalidB still validC partly validD unnecessaryShow answer & explanation →
Q18.
Mathematical induction is most useful for proving results involving a variable:
A n, a positive integerB x, a real numberC z, a complex numberD θ, an angleShow answer & explanation →
Q19.
The two essential parts of an induction proof are the base case and the:
A inductive stepB final stepC guessing stepD drawing stepShow answer & explanation →
Q20.
After both steps are proved, the statement is guaranteed for:
A all naturals from the base upB only for the value n = 1C only the even numbers involvedD for essentially no numbersShow answer & explanation →
Medium - 20 questions Q21.
In proving 1+2+...+n = n(n+1)/2 by induction, the inductive step adds which term to both sides of the assumed equation?
Show answer & explanation →
Q22.
After adding (k+1) to k(k+1)/2 in the induction proof of the natural number sum formula, the right-hand side simplifies to:
A (k+1)(k+2)/2B k(k+2)/2C (k+1)(k+1)/2D k(k+1)/2 + 1Show answer & explanation →
Q23.
To prove the sum of the first n odd numbers equals n<sup>2</sup>, what is the inductive hypothesis P(k)?
A 1+3+5+...+(2k-1) = k<sup>2</sup>B 1+3+5+...+(2k+1) = k<sup>2</sup>C 1+2+...+k = k<sup>2</sup>D k = k<sup>2</sup>Show answer & explanation →
Q24.
In the inductive step for the sum of first n odd numbers, what is added to k<sup>2</sup> to get the next term?
A 2k - 1B 2k + 1C k + 1D 2kShow answer & explanation →
Q25.
While proving n<sup>3</sup> - n is divisible by 6, (k+1)<sup>3</sup> - (k+1) is rewritten as (k<sup>3</sup> - k) plus which extra term?
A 3k<sup>2</sup> + 3kB 3k + 3C k<sup>2</sup> + kD 6kShow answer & explanation →
Q26.
Why is 3k(k+1) always divisible by 6 in the n<sup>3</sup> - n divisibility proof?
A Because k tends to be even in many natural number cases, independent of the value of k+1B Because k(k+1) is a product of consecutive integers, so it is always even, making 3k(k+1) divisible by 6C Because 3 itself is divisible by 6 without remainder, which some take as sufficient reasoning on its ownD Because k+1 happens to be divisible by 3 in this case, regardless of which natural number k is chosenShow answer & explanation →
Q27.
To prove 4<sup>n</sup> - 1 is divisible by 3 for all natural numbers n, the inductive step considers 4<sup>k+1</sup> - 1, which can be written as:
A 4(4<sup>k</sup> - 1) + 3B 4<sup>k</sup> - 1 + 4C 4 x 4<sup>k</sup> - 4D 4<sup>k</sup> + 3Show answer & explanation →
Q28.
For the inequality 2<sup>n</sup> > n, the inductive step shows 2<sup>k+1</sup> = 2 x 2<sup>k</sup> > 2k. To complete the proof we additionally need:
A 2k >= k+1 for k >= 1B k greater than 0, with no other condition neededC 2<sup>k</sup> less than k, an inequality that rarely holdsD k is an even natural number specificallyShow answer & explanation →
Q29.
Using induction to prove the sum of squares formula n(n+1)(2n+1)/6, what is P(2)?
A 1 + 4 = 2(3)(5)/6B 1 + 4 = 10C 5 = 5D 1 + 2 = 5Show answer & explanation →
Q30.
Which statement about induction is correct?
A Induction can prove a statement for just one specific value of n, leaving every other value of n unverifiedB Induction proves a statement for all n from the base case onward by chaining the inductive step infinitelyC Induction requires checking each natural number individually by hand, one at a time, indefinitelyD Induction works mainly for inequality statements, rarely for equalities between expressionsShow answer & explanation →
Q34.
In the inductive step for a sum formula, we add ___ to both sides of the assumed equation:
A the (k + 1)th termB the very first termC the number zeroD the value nShow answer & explanation →
Q37.
In an induction proof of divisibility, the (k + 1) case is handled using the:
A the inductive hypothesisB just the base caseC a final guessed valueD a random chosen numberShow answer & explanation →
Q38.
For the statement 2ⁿ > n, the base case n = 1 gives:
A 2 > 1, which is trueB 2 < 1, which is falseC 1 = 1 exactlyD 0 > 1 falselyShow answer & explanation →
Hard - 28 questions Q41.
To prove n! > 2<sup>n</sup> for all natural numbers n >= 4 by induction, what should the base case be?
A n = 1B n = 4C n = 0D n = 2Show answer & explanation →
Q42.
In the inductive step for n! > 2<sup>n</sup> (n >= 4), assuming k! > 2<sup>k</sup>, how is (k+1)! related to k!?
A (k+1)! = (k+1) x k!, and since k+1 > 2 for k >= 4, (k+1)! > 2 x k! > 2 x 2<sup>k</sup> = 2<sup>k+1</sup>B (k+1)! = k! + (k+1), treating factorial growth as additive rather than multiplicativeC (k+1)! = 2 x k!, fixing the multiplier at 2 regardless of the actual value of k+1D (k+1)! is unrelated to k!, since each factorial is computed independentlyShow answer & explanation →
Q43.
To prove that 7<sup>n</sup> - 3<sup>n</sup> is divisible by 4 for all natural numbers n, the inductive step writes 7<sup>k+1</sup> - 3<sup>k+1</sup> as:
A 7(7<sup>k</sup> - 3<sup>k</sup>) + 4(3<sup>k</sup>)B 7<sup>k</sup> - 3<sup>k</sup> + 4C 7 x 7<sup>k</sup> - 3 x 3<sup>k</sup> only, with no simplificationD 4(7<sup>k</sup> - 3<sup>k</sup>)Show answer & explanation →
Q44.
A student tries to prove P(n): n<sup>2</sup> = n by induction, checking P(1): 1 = 1 (true), and somehow forces an inductive step to look valid. What is the actual flaw?
A The base case itself is actually wrong here, since one squared does not truly equal one in this contextB P(n) is simply false for n=2 onward, so no valid inductive step can exist (any apparent proof has an algebraic error)C Induction as a method rarely applies cleanly to quadratic statements like n<sup>2</sup> = n in general cases like thisD The flaw lies mainly in how the final concluding sentence of the proof happens to be phrased, not in the underlying logicShow answer & explanation →
Q45.
To prove that 10<sup>n</sup> + 3 x 4<sup>n+2</sup> + 5 is divisible by 9 for all natural numbers n, what is the value at n=1, and is it divisible by 9?
A 10 + 192 + 5 = 207, and 207/9 = 23, so yesB 10 + 48 + 5 = 63, and 63/9 = 7, so yesC 207, but it is not divisible by 9D 10 + 3 + 5 = 18, divisible by 9Show answer & explanation →
Q46.
While proving the sum of cubes formula [n(n+1)/2]<sup>2</sup> by induction, the inductive step must show that [k(k+1)/2]<sup>2</sup> + (k+1)<sup>3</sup> equals:
A [(k+1)(k+2)/2]<sup>2</sup>B [k(k+2)/2]<sup>2</sup>C (k+1)<sup>2</sup> (k+2)<sup>2</sup>D k<sup>2</sup>(k+1)<sup>2</sup>/4 + (k+1)Show answer & explanation →
Q47.
Which of the following is the correct inductive hypothesis when proving that n(n+1)(n+2) is divisible by 6 for all natural numbers n?
A Assume k(k+1)(k+2) is divisible by 6, then show (k+1)(k+2)(k+3) is divisible by 6B Assume n itself is divisible by 6, without ever involving consecutive products togetherC Assume just the single term (k+1) is divisible by 6, ignoring the restD Assume k(k+1) is divisible by 3, dropping the third factor from considerationShow answer & explanation →
Q48.
For proving n<sup>2</sup> < 2<sup>n</sup> for all natural numbers n >= 5, what must be verified before applying the inductive step?
A That the base case n=5 holds: 25 < 32B That n=1 holds: 1 < 2C Nothing, induction applies automaticallyD That n=4 holds: 16 < 16Show answer & explanation →
Q49.
While proving 2<sup>2n</sup> - 1 is divisible by 3 by induction, the inductive step rewrites 2^(2(k+1)) - 1 as:
A (2<sup>2k</sup> - 1) + 4B 4(2<sup>2k</sup> - 1) + 3C 4(2<sup>2k</sup> - 1) - 3D 2(2<sup>2k</sup> - 1) + 1Show answer & explanation →
Q50.
A student proves P(k) implies P(k+1) for every k, but never verifies P(1). Even though the inductive step is logically correct, why does the proof fail?
A Without a verified base case, the chain of implications has no starting point to begin fromB The inductive hypothesis must always be assumed false, not true, for the argument to workC The statement P(n) must instead be proven separately for every even and odd value of nD Strong induction is required instead, since ordinary induction cannot use implicationsShow answer & explanation →
Q51.
To prove 3²ⁿ − 1 is divisible by 8, the base case n = 1 gives the value:
A 8, and 8 divides itB 9, not divisible by 8C 3, not divisible by 8D 0, trivially specialShow answer & explanation →
Q52.
In proving 1 + 2 + … + n = n(n + 1)/2, adding (k + 1) to k(k + 1)/2 gives:
A (k + 1)(k + 2)/2B only (k + 1)/2C only k(k + 1)/2D only (k + 2)/2Show answer & explanation →
Q54.
Which kind of result is proved most naturally by mathematical induction?
A a summation formula over n termsB the value of a single numberC a geometric construction resultD the value of a certain limitShow answer & explanation →
Q55.
If P(k) is assumed and P(k + 1) is proved, but the base case is false, then:
A the statement is not provenB the statement is provenC it holds for k onlyD it holds for all nShow answer & explanation →
Q56.
In proving (xⁿ − yⁿ) is divisible by (x − y), the (k + 1) case is rewritten using:
A that xᵏ − yᵏ is divisibleB only the given base caseC a purely lucky guessD no earlier known resultShow answer & explanation →
Q58.
The principle of mathematical induction is a fundamental property of the set of:
A natural numbersB real numbersC rational numbersD irrational numbersShow answer & explanation →
Q59.
Strong (complete) induction assumes the statement holds for:
A all values up to kB only the value kC only the value 1D no values at allShow answer & explanation →
Q60.
The most reliable way to prove that a summation formula holds for every positive integer is:
A mathematical inductionB checking a few casesC differentiating itD drawing a graphShow answer & explanation →