A statement $P(n)$ is true for $n = 1$, but the inductive step cannot be proved. Then

we cannot conclude that $P(n)$ is true for all $n$

$P(n)$ is true for all $n \ge 2$

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

$\frac{n (2 n^{2} + 7 n + 7)}{2}$

$\frac{(n - 1) (2 n^{2} + 3 n + 2)}{2}$

$\frac{(n + 1) (2 n^{2} + 11 n + 16)}{2}$

Maths 1A: Mathematical Induction: MCQ test, Set 8

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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.

Let $P(n)$ be $3 + 11 + 19 + \dots + (8 n - 5) = n (4 n - 1)$ . In $P(4)$ , the left side $3 + 11 + 19 + 27$ equals
A $95$ B $27$ C $33$ D $60$

Answer and solution

Answer: (D) $60$

$3 + 11 + 19 + 27 = 60$

and the right side at $n = 4$ is also $60$ .
A statement $P(n)$ is true for $n = 1$ , but the inductive step cannot be proved. Then
A $P(n)$ is true for all $n$ B $P(n)$ is false for all $n$ C we cannot conclude that $P(n)$ is true for all $n$ D $P(n)$ is true for all $n \ge 2$

Answer and solution

Answer: (C) we cannot conclude that $P(n)$ is true for all $n$

Both steps are needed. Without the inductive step, the principle gives no conclusion.
To prove $\displaystyle 1 + 9 + 81 + \dots + 9^{n - 1} = \frac{9^{n} - 1}{8}$ by induction, the term added to both sides of $P(k)$ is
A $9^{k - 1}$ B $9^{k}$ C $9^{k - 1} + 1$ D $9^{k + 1}$

Answer and solution

Answer: (B) $9^{k}$

The $(k+1)$ th term is got by putting $n = k + 1$ in $9^{n - 1}$ .

It is $9^{k}$ .
By induction it can be shown that $\displaystyle \frac{1}{1 \cdot 7} + \frac{1}{7 \cdot 13} + \frac{1}{13 \cdot 19} + \dots + \frac{1}{(6 n - 5)(6 n + 1)} =$
A $\displaystyle \frac{2 n}{6 n + 1}$ B $\displaystyle \frac{n - 1}{6 n - 5}$ C $\displaystyle \frac{n + 1}{6 n + 7}$ D $\displaystyle \frac{n}{6 n + 1}$

Answer and solution

Answer: (D) $\displaystyle \frac{n}{6 n + 1}$

Check with $n = 1$ : the sum is $\displaystyle \frac{1}{7}$ and the formula gives $\displaystyle \frac{1}{7}$ .

The formula $\displaystyle \frac{n}{6 n + 1}$ is the one proved by induction.
If $1 + 2 + 3 + \dots + n = 276$ , then $n =$
A $22$ B $46$ C $24$ D $23$

Answer and solution

Answer: (D) $23$

$\displaystyle \frac{n(n+1)}{2} = 276 \Rightarrow n(n+1) = 552 = 23 \cdot 24$

$\therefore n = 23$
$1^2 + 2^2 + \dots + 17^2 =$
A $23409$ B $3570$ C $1785$ D $1683$

Answer and solution

Answer: (C) $1785$

$\displaystyle \frac{n(n+1)(2n+1)}{6} = \frac{17 \cdot 18 \cdot 35}{6} = 1785$
$2^2 + 4^2 + \dots + 22^2 =$
A $2024$ B $3795$ C $1012$ D $506$

Answer and solution

Answer: (A) $2024$

$\displaystyle 4\sum_{k=1}^{11} k^2 = 4 \cdot 506 = 2024$
$1^3 + 3^3 + 5^3 + \dots$ to $9$ terms $=$
A $1377$ B $13203$ C $2025$ D $13041$

Answer and solution

Answer: (D) $13041$

$n^2(2n^2 - 1) = 81(161) = 13041$
$1^2 - 2^2 + 3^2 - \dots - 30^2 =$
A $-930$ B $465$ C $435$ D $-465$

Answer and solution

Answer: (D) $-465$

$n = 30$ is even.

$\displaystyle -\frac{30 \cdot 31}{2} = -465$
$\displaystyle \sum_{k=1}^{n} (k + 1) (3 k + 1) =$
A $\displaystyle \frac{n (2 n^{2} + 7 n + 7)}{2}$ B $n (2 n^{2} + 7 n + 7)$ C $\displaystyle \frac{(n - 1) (2 n^{2} + 3 n + 2)}{2}$ D $\displaystyle \frac{(n + 1) (2 n^{2} + 11 n + 16)}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{n (2 n^{2} + 7 n + 7)}{2}$

Expand $(k + 1) (3 k + 1) = 3 k^{2} + 4 k + 1$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (2 n^{2} + 7 n + 7)}{2}$ .
$\displaystyle \sum_{k=1}^{5} k (k + 1) (k + 2) =$
A $1050$ B $420$ C $756$ D $210$

Answer and solution

Answer: (B) $420$

Expand $k (k + 1) (k + 2) = k^{3} + 3 k^{2} + 2 k$ and use the standard sums with $n = 5$ .

The sum is $420$ .
The sum of the first $26$ terms of the A.P. $9, 11, 13, \dots$ is
A $910$ B $884$ C $767$ D $1534$

Answer and solution

Answer: (B) $884$

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{26}{2}[18 + 50] = 884$
The sum of the first $8$ terms of $5 + 10 + 20 + \dots$ is
A $2555$ B $635$ C $1275$ D $1280$

Answer and solution

Answer: (C) $1275$

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 5 \cdot \frac{256 - 1}{1} = 1275$
The sum to infinity of $\displaystyle 15 - 10 + \frac{20}{3} - \frac{40}{9} + \dots$ is
A $25$ B $45$ C $9$ D $\displaystyle - \frac{45}{2}$

Answer and solution

Answer: (C) $9$

$a = 15$ , $\displaystyle r = - \frac{2}{3}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = 9$
The sum to infinity of $\displaystyle 5 + 6 \cdot \frac{1}{4} + 7 \cdot \frac{1}{16} + 8 \cdot \frac{1}{64} + \dots$ is
A $\displaystyle \frac{64}{9}$ B $7$ C $\displaystyle \frac{20}{3}$ D $\displaystyle \frac{76}{9}$

Answer and solution

Answer: (A) $\displaystyle \frac{64}{9}$

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

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{20}{3} + \frac{4}{9} = \frac{64}{9}$
The $n$ th term of $8, 11, 16, 23, \dots$ is
A $2 n^{2} - n + 6$ B $n^{2} + 7$ C $n^{2}$ D $n^{2} + 2 n + 8$

Answer and solution

Answer: (B) $n^{2} + 7$

Differences: $3, 5, 7$ .

Fitting $n = 1, 2, 3$ gives $T_n = n^{2} + 7$ .
The $n$ th term of $11, 51, 251, 1251, \dots$ is
A $2 \cdot 5^{n + 1} + 1$ B $2 \cdot 5^{n}$ C $2 \cdot 5^{n} + 1$ D $2 \cdot 5^{n} - 1$

Answer and solution

Answer: (C) $2 \cdot 5^{n} + 1$

Differences: $40, 200, 1000$ .

Fitting $n = 1, 2, 3$ gives $T_n = 2 \cdot 5^{n} + 1$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $5 + 17 + 29 + \dots$ is
A $12 n - 7$ B $12 n + 5$ C $13 n - 8$ D $12 n - 19$

Answer and solution

Answer: (A) $12 n - 7$

$T_1 = 5$ , $T_2 = 17$ .

Only $12 n - 7$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $4 + 7 + 10 + \dots$ is
A $\displaystyle \frac{n (3 n + 5)}{2}$ B $\displaystyle \frac{3 n^{2} + 7 n - 2}{2}$ C $n (3 n + 1)$ D $\displaystyle \frac{3 n^{2} - n + 6}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{n (3 n + 5)}{2}$

$S_1 = 4$ , $S_2 = 11$ , $S_3 = 21$ .

Only $\displaystyle \frac{n (3 n + 5)}{2}$ gives all three.
The largest number that divides $4^{2 n} - 1$ for every $n \in \mathbb{N}$ is
A $30$ B $15$ C $45$ D $16$

Answer and solution

Answer: (B) $15$

$n = 1:\ 15,\ n = 2:\ 255$

$\text{G.C.D.}(15, 255) = 15$
The largest number that divides $6^{n} - 1^{n}$ for every $n \in \mathbb{N}$ is
A $5$ B $6$ C $10$ D $15$

Answer and solution

Answer: (A) $5$

$n = 1:\ 5,\ n = 2:\ 35$

$\text{G.C.D.}(5, 35) = 5$
To prove $15^n - 4^n$ is divisible by $11$ , assume $15^k - 4^k = 11\lambda$ . Then $15^{k+1} - 4^{k+1} =$
A $11(15\lambda - 4^k)$ B $11(15\lambda + 4^k)$ C $11(4\lambda + 15^k)$ D $11(\lambda + 4^k)$

Answer and solution

Answer: (B) $11(15\lambda + 4^k)$

$15^{k+1} - 4^{k+1} = 15(11\lambda + 4^k) - 4 \cdot 4^k$

$= 165\lambda + 11 \cdot 4^k = 11(15\lambda + 4^k)$
With $P = 5$ , $P^{n+1} + (P + 1)^{2n-1}$ is divisible by
A $30$ B $31$ C $11$ D $36$

Answer and solution

Answer: (B) $31$

$P^2 + P + 1 = 25 + 5 + 1 = 31$ .
If $137 \equiv r \pmod{15}$ with $0 \le r < 15$ , then $r =$
A $13$ B $9$ C $3$ D $2$

Answer and solution

Answer: (D) $2$

$137 = 15 \times 9 + 2$

$\therefore 137 - 2 = 135$ is divisible by $15$ , so $r = 2$ .
If $82 \equiv r \pmod{9}$ with $0 \le r < 9$ , then $r =$
A $2$ B $8$ C $9$ D $1$

Answer and solution

Answer: (D) $1$

$82 = 9 \times 9 + 1$

$\therefore 82 - 1 = 81$ is divisible by $9$ , so $r = 1$ .

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MCQ test, Set 8