In the principle of complete induction, to prove $P(k+1)$ we may assume that

$P(1), P(2), \dots, P(k)$ are all true

$\sum_{k=1}^{n} k (2 k + 3) =$

$\frac{n (n + 1) (4 n + 11)}{6}$

$\frac{n (n - 1) (4 n + 7)}{6}$

$\frac{n (n + 1) (4 n + 11)}{3}$

$\frac{(n + 1) (n + 2) (4 n + 15)}{6}$

Maths 1A: Mathematical Induction: MCQ test, Set 3

Download the PDF (with key sheet and solutions, 86.6 KB)

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 $P(n)$ : $5 + 9 + 13 + \dots + (4 n + 1) = n (2 n + 3)$ , then the value of each side of $P(4)$ is
A $44$ B $27$ C $17$ D $65$

Answer and solution

Answer: (A) $44$

$5 + 9 + 13 + 17 = 44$

and the right side at $n = 4$ is also $44$ .
In the principle of complete induction, to prove $P(k+1)$ we may assume that
A only $P(1)$ is true B $P(1), P(2), \dots, P(k)$ are all true C $P(k+2)$ is true D $P(k+1)$ is true

Answer and solution

Answer: (B) $P(1), P(2), \dots, P(k)$ are all true

Complete induction lets us use every earlier case, $P(1)$ to $P(k)$ .
To prove $\displaystyle 3^2 + 5^2 + 7^2 + \dots + (2 n + 1)^{2} = \frac{n (4 n^{2} + 12 n + 11)}{3}$ by induction, the term added to both sides of $P(k)$ is
A $(2 k + 3)^{2}$ B $(2 k + 1)^{2}$ C $(2 k + 5)^{2}$ D $(2 k + 1)^{2} + 1$

Answer and solution

Answer: (A) $(2 k + 3)^{2}$

The $(k+1)$ th term is got by putting $n = k + 1$ in $(2 n + 1)^{2}$ .

It is $(2 k + 3)^{2}$ .
By induction it can be shown that $6 + 13 + 20 + \dots + (7 n - 1) =$
A $\displaystyle \frac{(n - 1) (7 n - 2)}{2}$ B $n (7 n + 5)$ C $\displaystyle \frac{(n + 1) (7 n + 12)}{2}$ D $\displaystyle \frac{n (7 n + 5)}{2}$

Answer and solution

Answer: (D) $\displaystyle \frac{n (7 n + 5)}{2}$

Check with $n = 1$ : the sum is $6$ and the formula gives $6$ .

The formula $\displaystyle \frac{n (7 n + 5)}{2}$ is the one proved by induction.
If $1 + 2 + 3 + \dots + n = 136$ , then $n =$
A $32$ B $15$ C $16$ D $17$

Answer and solution

Answer: (C) $16$

$\displaystyle \frac{n(n+1)}{2} = 136 \Rightarrow n(n+1) = 272 = 16 \cdot 17$

$\therefore n = 16$
$1^2 + 2^2 + \dots + 14^2 =$
A $11025$ B $1015$ C $945$ D $2030$

Answer and solution

Answer: (B) $1015$

$\displaystyle \frac{n(n+1)(2n+1)}{6} = \frac{14 \cdot 15 \cdot 29}{6} = 1015$
$1^3 + 3^3 + 5^3 + \dots$ to $5$ terms $=$
A $2025$ B $225$ C $1275$ D $1225$

Answer and solution

Answer: (D) $1225$

$n^2(2n^2 - 1) = 25(49) = 1225$
$1^2 + 3^2 + 5^2 + \dots$ to $n$ terms $=$
A $n^2$ B $\displaystyle \frac{n}{3}(4n^2 - 1)$ C $\displaystyle \frac{n}{6}(4n^2 - 1)$ D $\displaystyle \frac{n}{3}(4n^2 + 1)$

Answer and solution

Answer: (B) $\displaystyle \frac{n}{3}(4n^2 - 1)$

The standard result for the squares of odd numbers.

$\displaystyle \sum (2k-1)^2 = \frac{n}{3}(4n^2 - 1)$
$1^2 - 2^2 + 3^2 - 4^2 + \dots$ to $2n$ terms $=$
A $\displaystyle -\frac{n(n+1)}{2}$ B $-n(2n+1)$ C $n(2n+1)$ D $-2n(2n+1)$

Answer and solution

Answer: (B) $-n(2n+1)$

$2n$ is even, so the sum is $\displaystyle -\frac{2n(2n+1)}{2} = -n(2n+1)$ .
$\displaystyle \sum_{k=1}^{n} k (2 k + 3) =$
A $\displaystyle \frac{n (n - 1) (4 n + 7)}{6}$ B $\displaystyle \frac{n (n + 1) (4 n + 11)}{3}$ C $\displaystyle \frac{n (n + 1) (4 n + 11)}{6}$ D $\displaystyle \frac{(n + 1) (n + 2) (4 n + 15)}{6}$

Answer and solution

Answer: (C) $\displaystyle \frac{n (n + 1) (4 n + 11)}{6}$

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

The total factorises as $\displaystyle \frac{n (n + 1) (4 n + 11)}{6}$ .
$\displaystyle \sum_{k=1}^{7} k^{2} + 2 k =$
A $441$ B $196$ C $133$ D $276$

Answer and solution

Answer: (B) $196$

Expand $k^{2} + 2 k = k^{2} + 2 k$ and use the standard sums with $n = 7$ .

The sum is $196$ .
The $31$ th term of the A.P. $10, 14, 18, \dots$ is
A $130$ B $126$ C $134$ D $124$

Answer and solution

Answer: (A) $130$

$t_n = a + (n-1)d = 10 + 30(4) = 130$
The $9$ th term of the G.P. $5, 10, 20, \dots$ is
A $100000000$ B $2560$ C $640$ D $1280$

Answer and solution

Answer: (D) $1280$

$t_n = a r^{n-1} = 5 \cdot 2^{8} = 1280$
The sum of the first $7$ terms of $1 + 3 + 9 + \dots$ is
A $2187$ B $2186$ C $3280$ D $1093$

Answer and solution

Answer: (D) $1093$

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 1 \cdot \frac{2187 - 1}{2} = 1093$
The sum to infinity of $\displaystyle 1 + 6 \cdot \frac{1}{4} + 11 \cdot \frac{1}{16} + 16 \cdot \frac{1}{64} + \dots$ is
A $\displaystyle \frac{4}{3}$ B $\displaystyle \frac{32}{9}$ C $3$ D $\displaystyle \frac{92}{9}$

Answer and solution

Answer: (B) $\displaystyle \frac{32}{9}$

$a = 1$ , $d = 5$ , $\displaystyle r = \frac{1}{4}$ .

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{4}{3} + \frac{20}{9} = \frac{32}{9}$
The $n$ th term of $5, 6, 9, 14, \dots$ is
A $n^{2} + 5$ B $n^{2} - 2 n$ C $n^{2} - 2 n + 6$ D $2 n^{2} - 3 n + 5$

Answer and solution

Answer: (C) $n^{2} - 2 n + 6$

Differences: $1, 3, 5$ .

Fitting $n = 1, 2, 3$ gives $T_n = n^{2} - 2 n + 6$ .
The $n$ th term of $8, 28, 128, 628, \dots$ is
A $5^{n} - 3$ B $5^{n + 1} + 3$ C $5^{n}$ D $5^{n} + 3$

Answer and solution

Answer: (D) $5^{n} + 3$

Differences: $20, 100, 500$ .

Fitting $n = 1, 2, 3$ gives $T_n = 5^{n} + 3$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $2 \cdot 4 + 4 \cdot 6 + 6 \cdot 8 + \dots$ is
A $4 n (n - 1)$ B $2 (n + 1) (2 n + 4)$ C $2 n (2 n + 2)$ D $4 n^{2} + 5 n - 1$

Answer and solution

Answer: (C) $2 n (2 n + 2)$

$T_1 = 8$ , $T_2 = 24$ .

Only $2 n (2 n + 2)$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $1 + 4 + 7 + \dots$ is
A $\displaystyle \frac{n (3 n - 1)}{2}$ B $\displaystyle \frac{3 n^{2} - 7 n + 6}{2}$ C $n (3 n - 2)$ D $\displaystyle \frac{(n + 1) (3 n - 2)}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{n (3 n - 1)}{2}$

$S_1 = 1$ , $S_2 = 5$ , $S_3 = 12$ .

Only $\displaystyle \frac{n (3 n - 1)}{2}$ gives all three.
The largest number that divides $10^{n} - 1$ for every $n \in \mathbb{N}$ is
A $10$ B $18$ C $27$ D $9$

Answer and solution

Answer: (D) $9$

$n = 1:\ 9,\ n = 2:\ 99$

$\text{G.C.D.}(9, 99) = 9$
The largest number that divides $2^{2 n} - 1$ for every $n \in \mathbb{N}$ is
A $4$ B $9$ C $6$ D $3$

Answer and solution

Answer: (D) $3$

$n = 1:\ 3,\ n = 2:\ 15$

$\text{G.C.D.}(3, 15) = 3$
To prove $13^n - 4^n$ is divisible by $9$ , assume $13^k - 4^k = 9\lambda$ . Then $13^{k+1} - 4^{k+1} =$
A $9(4\lambda + 13^k)$ B $9(13\lambda - 4^k)$ C $9(13\lambda + 4^k)$ D $9(\lambda + 4^k)$

Answer and solution

Answer: (C) $9(13\lambda + 4^k)$

$13^{k+1} - 4^{k+1} = 13(9\lambda + 4^k) - 4 \cdot 4^k$

$= 117\lambda + 9 \cdot 4^k = 9(13\lambda + 4^k)$
If $65 \equiv r \pmod{9}$ with $0 \le r < 9$ , then $r =$
A $2$ B $4$ C $3$ D $7$

Answer and solution

Answer: (A) $2$

$65 = 9 \times 7 + 2$

$\therefore 65 - 2 = 63$ is divisible by $9$ , so $r = 2$ .
The sum of the cubes of three consecutive natural numbers is always divisible by
A $27$ B $8$ C $9$ D $6$

Answer and solution

Answer: (C) $9$

Standard result: divisible by $9$ .

For example $1 + 8 + 27 = 36 = 9 \times 4$ .
If $94 \equiv r \pmod{11}$ with $0 \le r < 11$ , then $r =$
A $5$ B $7$ C $8$ D $6$

Answer and solution

Answer: (D) $6$

$94 = 11 \times 8 + 6$

$\therefore 94 - 6 = 88$ is divisible by $11$ , so $r = 6$ .

Menu

MCQ test, Set 3