Module 5 Module Quiz Extension 1

Module 5 Quiz

15 multiple choice questions across the whole module: the structure of induction proofs, series results, divisibility, inequalities, recurrence relations, and spotting flawed arguments.

Module 5 Quiz

Q11 MARK

Which statement is the usual base case for proving $1 + 2 + ... + n = n(n + 1)/2$ for all positive integers $n$?

ACheck $n = 0$ BCheck $n = 1$ CAssume the result for $n = k$ DCheck $n = k + 1$

Q21 MARK

In an induction proof, the induction hypothesis is the statement that

Athe result is true for every real number Bthe result is false for $k + 1$ Cthe base case can be skipped Dthe result is true for a particular integer $k$

Q31 MARK

To prove a divisibility result such as $3$ divides $4^{n} - 1$, the induction step should usually show

A$4^{k} - 1$ is prime B$4^{k + 1} - 1$ is less than $4^{k} - 1$ Cthe result only works for even $k$ D$4^{k + 1} - 1$ can be written using $4^{k} - 1$ plus a multiple of $3$

Q41 MARK

If a proof assumes $P(k)$ and then proves $P(k + 2)$, what extra condition is usually needed for all integers from the start value?

AA contradiction step BA derivative test CTwo consecutive base cases DA change to decimal form

Q51 MARK

Which algebraic move is most useful in proving a summation formula by induction?

AReplace the sum to $k$ using the induction hypothesis, then add the $(k + 1)$th term BDifferentiate every term CSet $k = -1$ DIgnore the final term

Q61 MARK

A common flaw in induction proofs is circular reasoning. This happens when the proof

Achecks a base case Bassumes the statement for $k$ Cuses the statement for $k + 1$ before it has been proved Dsimplifies both sides

Q71 MARK

Strong induction differs from ordinary induction because the induction hypothesis may assume

Aonly $P(k + 1)$ Bno earlier cases Cthe result for all real $x$ D$P(1), P(2), ..., P(k)$

Q81 MARK

When proving an inequality by induction, which issue most often needs careful attention?

AWhether the variable is written in radians BWhether multiplying or adding terms preserves the inequality direction CWhether the expression has a graph DWhether the coefficient is complex

Q91 MARK

For a recurrence-defined sequence, an induction proof is often suitable because

Aeach term is defined using earlier terms Ball sequences are arithmetic Crecurrences never require base cases Dthe variable is continuous

Q101 MARK

Which is the clearest final sentence in an induction proof?

AThis seems true for many values. BTherefore it is true for $k$. CSince the base case is true and $P(k)$ implies $P(k + 1)$, the result holds for all required integers. DThe answer is probably correct.

Q111 MARK

To prove a product formula by induction, $P(k + 1)$ is usually obtained by

Asubtracting the first factor Btaking the logarithm only Creplacing the product to $k$ using $P(k)$, then multiplying by the next factor Dtesting one large value

Q121 MARK

If the required statement begins at $n = 4$, the base case should be

A$n = 0$ B$n = 1$ C$n = 3$ D$n = 4$

Q131 MARK

Which expression is equivalent to $2^{k + 1} - 1$ in a form that uses $2^{k} - 1$?

A$2(2^{k} - 1) + 1$ B$2^{k} - 2$ C$2^{k} + 1$ D$2(2^{k} + 1)$

Q141 MARK

In a proof by contradiction, the first step is to

Astate the desired conclusion twice Bassume the negation of the desired conclusion Cperform induction without a base case Dchoose a random counterexample

Q151 MARK

Which situation is best suited to induction?

AA single numerical evaluation BA statement about all positive integers CA limit as $x$ approaches $0$ DA sketch of a continuous graph