Strong induction is another name for

the principle of complete induction

the principle of finite induction

$1^3 + 2^3 + 3^3 + \dots + n^3$ is equal to

$\left(\frac{n(n+1)}{2}\right)^2$

$x \equiv y \pmod{m}$ means

$x$ and $y$ are both divisible by $m$

Maths 1A: Mathematical Induction: MCQ test, Set 5

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Choose one answer for each question, then check your score at the end. Each correct answer carries 1 mark; there is no negative marking.

If $P(n)$ : $\displaystyle 7 + 10 + 13 + \dots + (3 n + 4) = \frac{n (3 n + 11)}{2}$ , then the value of each side of $P(4)$ is
A $46$ B $30$ C $16$ D $65$

Answer and solution

Answer: (A) $46$

$7 + 10 + 13 + 16 = 46$

and the right side at $n = 4$ is also $46$ .
Strong induction is another name for
A the principle of complete induction B the induction hypothesis C the principle of finite induction D the base step

Answer and solution

Answer: (A) the principle of complete induction

The principle of complete induction is also called strong induction.
To prove $\displaystyle 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 + \dots + n(n + 3) = \frac{n (n + 1) (n + 5)}{3}$ by induction, the term added to both sides of $P(k)$ is
A $k (k + 3) + 1$ B $(k + 2) (k + 5)$ C $(k + 1) (k + 4)$ D $k (k + 3)$

Answer and solution

Answer: (C) $(k + 1) (k + 4)$

The $(k+1)$ th term is got by putting $n = k + 1$ in $n(n + 3)$ .

It is $(k + 1) (k + 4)$ .
By induction it can be shown that $\displaystyle \frac{1}{1 \cdot 6} + \frac{1}{6 \cdot 11} + \frac{1}{11 \cdot 16} + \dots + \frac{1}{(5 n - 4)(5 n + 1)} =$
A $\displaystyle \frac{n - 1}{5 n - 4}$ B $\displaystyle \frac{n}{5 n + 1}$ C $\displaystyle \frac{n + 1}{5 n + 6}$ D $\displaystyle \frac{2 n}{5 n + 1}$

Answer and solution

Answer: (B) $\displaystyle \frac{n}{5 n + 1}$

Check with $n = 1$ : the sum is $\displaystyle \frac{1}{6}$ and the formula gives $\displaystyle \frac{1}{6}$ .

The formula $\displaystyle \frac{n}{5 n + 1}$ is the one proved by induction.
If $1^3 + 2^3 + \dots + n^3 = 6084$ , then $1 + 2 + \dots + n =$
A $78$ B $12$ C $90$ D $3042$

Answer and solution

Answer: (A) $78$

$\displaystyle \sum n^3 = \left(\sum n\right)^2$

$\displaystyle \therefore \sum n = \sqrt{6084} = 78$
$1^3 + 2^3 + 3^3 + \dots + n^3$ is equal to
A $\displaystyle \frac{n(n+1)^2}{2}$ B $\displaystyle \left(\frac{n(n+1)}{2}\right)^2$ C $\displaystyle \frac{n^2(n+1)}{4}$ D $\displaystyle \frac{n(n+1)(2n+1)}{6}$

Answer and solution

Answer: (B) $\displaystyle \left(\frac{n(n+1)}{2}\right)^2$

The sum of the cubes is the square of the sum: $\displaystyle \left(\sum n\right)^2 = \frac{n^2(n+1)^2}{4}$ .
$1^2 + 3^2 + 5^2 + \dots$ to $11$ terms $=$
A $3311$ B $1793$ C $2024$ D $1771$

Answer and solution

Answer: (D) $1771$

$\displaystyle \frac{n}{3}(4n^2 - 1) = \frac{11}{3}(483) = 1771$
$2^2 + 4^2 + \dots + 24^2 =$
A $4900$ B $2600$ C $650$ D $1300$

Answer and solution

Answer: (B) $2600$

$\displaystyle 4\sum_{k=1}^{12} k^2 = 4 \cdot 650 = 2600$
$1^2 - 2^2 + 3^2 - \dots - 24^2 =$
A $-300$ B $276$ C $-600$ D $300$

Answer and solution

Answer: (A) $-300$

$n = 24$ is even.

$\displaystyle -\frac{24 \cdot 25}{2} = -300$
$\displaystyle \sum_{k=1}^{n} k (3 k - 1) =$
A $n (n - 1)^{2}$ B $n^{2} (n + 1)$ C $(n + 1)^{2} (n + 2)$ D $2 n^{2} (n + 1)$

Answer and solution

Answer: (B) $n^{2} (n + 1)$

Expand $k (3 k - 1) = 3 k^{2} - k$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $n^{2} (n + 1)$ .
$\displaystyle \sum_{k=1}^{5} k (2 k + 1) =$
A $275$ B $70$ C $203$ D $125$

Answer and solution

Answer: (D) $125$

Expand $k (2 k + 1) = 2 k^{2} + k$ and use the standard sums with $n = 5$ .

The sum is $125$ .
The $17$ th term of the A.P. $5, 11, 17, \dots$ is
A $101$ B $102$ C $107$ D $95$

Answer and solution

Answer: (A) $101$

$t_n = a + (n-1)d = 5 + 16(6) = 101$
The sum of the first $9$ terms of $4 + 8 + 16 + \dots$ is
A $4092$ B $1020$ C $2044$ D $2048$

Answer and solution

Answer: (C) $2044$

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 4 \cdot \frac{512 - 1}{1} = 2044$
The sum to infinity of $\displaystyle 10 + 6 + \frac{18}{5} + \frac{54}{25} + \dots$ is
A $25$ B $\displaystyle \frac{50}{3}$ C $\displaystyle \frac{25}{4}$ D $4$

Answer and solution

Answer: (A) $25$

$a = 10$ , $\displaystyle r = \frac{3}{5}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = 25$
The sum to infinity of $\displaystyle 4 + 7 \cdot \frac{1}{2} + 10 \cdot \frac{1}{4} + 13 \cdot \frac{1}{8} + \dots$ is
A $20$ B $11$ C $14$ D $8$

Answer and solution

Answer: (C) $14$

$a = 4$ , $d = 3$ , $\displaystyle r = \frac{1}{2}$ .

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = 8 + 6 = 14$
The $n$ th term of $1, 8, 19, 34, \dots$ is
A $2 n^{2} + 5 n + 1$ B $2 n^{2} + n$ C $3 n^{2} - 3$ D $2 n^{2} + n - 2$

Answer and solution

Answer: (D) $2 n^{2} + n - 2$

Differences: $7, 11, 15$ .

Fitting $n = 1, 2, 3$ gives $T_n = 2 n^{2} + n - 2$ .
The $n$ th term of $2, 14, 62, 254, \dots$ is
A $4^{n} + 2$ B $4^{n + 1} - 2$ C $4^{n}$ D $4^{n} - 2$

Answer and solution

Answer: (D) $4^{n} - 2$

Differences: $12, 48, 192$ .

Fitting $n = 1, 2, 3$ gives $T_n = 4^{n} - 2$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $2 \cdot 4 + 3 \cdot 5 + 4 \cdot 6 + \dots$ is
A $(n + 2) (n + 4)$ B $n (n + 2)$ C $(n + 1) (n + 3)$ D $n^{2} + 5 n + 2$

Answer and solution

Answer: (C) $(n + 1) (n + 3)$

$T_1 = 8$ , $T_2 = 15$ .

Only $(n + 1) (n + 3)$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $3 + 7 + 11 + \dots$ is
A $n (2 n + 1)$ B $2 n^{2} + 2 n - 1$ C $n (4 n - 1)$ D $2 n^{2} - 3 n + 4$

Answer and solution

Answer: (A) $n (2 n + 1)$

$S_1 = 3$ , $S_2 = 10$ , $S_3 = 21$ .

Only $n (2 n + 1)$ gives all three.
The largest number that divides $9^{n} - 1$ for every $n \in \mathbb{N}$ is
A $40$ B $16$ C $9$ D $8$

Answer and solution

Answer: (D) $8$

$n = 1:\ 8,\ n = 2:\ 80$

$\text{G.C.D.}(8, 80) = 8$
The largest number that divides $14^{n} - 9^{n}$ for every $n \in \mathbb{N}$ is
A $5$ B $10$ C $15$ D $6$

Answer and solution

Answer: (A) $5$

$n = 1:\ 5,\ n = 2:\ 115$

$\text{G.C.D.}(5, 115) = 5$
To prove $12^n - 5^n$ is divisible by $7$ , assume $12^k - 5^k = 7\lambda$ . Then $12^{k+1} - 5^{k+1} =$
A $7(5\lambda + 12^k)$ B $7(\lambda + 5^k)$ C $7(12\lambda - 5^k)$ D $7(12\lambda + 5^k)$

Answer and solution

Answer: (D) $7(12\lambda + 5^k)$

$12^{k+1} - 5^{k+1} = 12(7\lambda + 5^k) - 5 \cdot 5^k$

$= 84\lambda + 7 \cdot 5^k = 7(12\lambda + 5^k)$
If $88 \equiv r \pmod{13}$ with $0 \le r < 13$ , then $r =$
A $3$ B $11$ C $10$ D $6$

Answer and solution

Answer: (C) $10$

$88 = 13 \times 6 + 10$

$\therefore 88 - 10 = 78$ is divisible by $13$ , so $r = 10$ .
$x \equiv y \pmod{m}$ means
A $x$ and $y$ are both divisible by $m$ B $xy$ is divisible by $m$ C $x - y$ is divisible by $m$ D $x + y$ is divisible by $m$

Answer and solution

Answer: (C) $x - y$ is divisible by $m$

This is the definition of congruence modulo $m$ .
If $71 \equiv r \pmod{4}$ with $0 \le r < 4$ , then $r =$
A $1$ B $3$ C $17$ D $0$

Answer and solution

Answer: (B) $3$

$71 = 4 \times 17 + 3$

$\therefore 71 - 3 = 68$ is divisible by $4$ , so $r = 3$ .

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MCQ test, Set 5