In the domino picture of induction, the inductive step says that

each falling domino knocks over the next one

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

$\frac{n (n - 1) (2 n + 11)}{6}$

$\frac{n (n + 1) (2 n + 13)}{6}$

$\frac{(n - 2) (n - 1) (2 n + 9)}{6}$

$\frac{n (n - 1) (2 n + 11)}{3}$

Maths 1A: Mathematical Induction: MCQ test, Set 6

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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 $\displaystyle 1 + 6 + 11 + \dots + (5 n - 4) = \frac{n (5 n - 3)}{2}$ . In $P(2)$ , the left side $1 + 6$ equals
A $6$ B $18$ C $1$ D $7$

Answer and solution

Answer: (D) $7$

$1 + 6 = 7$

and the right side at $n = 2$ is also $7$ .
In the domino picture of induction, the inductive step says that
A the first domino falls B the last domino falls first C each falling domino knocks over the next one D all the dominoes fall at once

Answer and solution

Answer: (C) each falling domino knocks over the next one

The base step knocks over the first domino.

The inductive step: each falling domino knocks over the next.
To prove $\displaystyle 8 + 11 + 14 + \dots + (3 n + 5) = \frac{n (3 n + 13)}{2}$ by induction, the term added to both sides of $P(k)$ is
A $3 k + 6$ B $3 k + 8$ C $3 k + 5$ D $3 k + 11$

Answer and solution

Answer: (B) $3 k + 8$

The $(k+1)$ th term is got by putting $n = k + 1$ in $(3 n + 5)$ .

It is $3 k + 8$ .
By induction it can be shown that $7 + 16 + 25 + \dots + (9 n - 2) =$
A $n (9 n + 5)$ B $\displaystyle \frac{(n + 1) (9 n + 14)}{2}$ C $\displaystyle \frac{n (9 n + 5)}{2}$ D $\displaystyle \frac{(n - 1) (9 n - 4)}{2}$

Answer and solution

Answer: (C) $\displaystyle \frac{n (9 n + 5)}{2}$

Check with $n = 1$ : the sum is $7$ and the formula gives $7$ .

The formula $\displaystyle \frac{n (9 n + 5)}{2}$ is the one proved by induction.
$1^2 + 2^2 + \dots + 20^2 =$
A $2730$ B $44100$ C $5740$ D $2870$

Answer and solution

Answer: (D) $2870$

$\displaystyle \frac{n(n+1)(2n+1)}{6} = \frac{20 \cdot 21 \cdot 41}{6} = 2870$
$1^4 + 2^4 + \dots + 9^4 =$
A $81225$ B $4050$ C $15333$ D $8772$

Answer and solution

Answer: (C) $15333$

$\displaystyle \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ with $n = 9$

$= 15333$
$1^3 + 3^3 + 5^3 + \dots$ to $8$ terms $=$
A $1296$ B $8128$ C $960$ D $8256$

Answer and solution

Answer: (B) $8128$

$n^2(2n^2 - 1) = 64(127) = 8128$
The sum of the first $52$ odd natural numbers is
A $2756$ B $2704$ C $2703$ D $2809$

Answer and solution

Answer: (B) $2704$

$1 + 3 + \dots + (2n - 1) = n^2 = 2704$
$1^2 - 2^2 + 3^2 - \dots + 17^2 =$
A $153$ B $-153$ C $306$ D $-136$

Answer and solution

Answer: (A) $153$

$n = 17$ is odd.

$\displaystyle \frac{17 \cdot 18}{2} = 153$
$\displaystyle \sum_{k=1}^{n} (k - 1) (k + 3) =$
A $\displaystyle \frac{n (n - 1) (2 n + 11)}{6}$ B $\displaystyle \frac{n (n + 1) (2 n + 13)}{6}$ C $\displaystyle \frac{(n - 2) (n - 1) (2 n + 9)}{6}$ D $\displaystyle \frac{n (n - 1) (2 n + 11)}{3}$

Answer and solution

Answer: (A) $\displaystyle \frac{n (n - 1) (2 n + 11)}{6}$

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

The total factorises as $\displaystyle \frac{n (n - 1) (2 n + 11)}{6}$ .
$\displaystyle \sum_{k=1}^{8} 3 k^{2} - 1 =$
A $846$ B $413$ C $1528$ D $604$

Answer and solution

Answer: (D) $604$

Expand $3 k^{2} - 1 = 3 k^{2} - 1$ and use the standard sums with $n = 8$ .

The sum is $604$ .
The sum of the first $20$ terms of the A.P. $12, 15, 18, \dots$ is
A $690$ B $840$ C $810$ D $1380$

Answer and solution

Answer: (C) $810$

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{20}{2}[24 + 57] = 810$
The $6$ th term of the G.P. $7, 14, 28, \dots$ is
A $112$ B $537824$ C $224$ D $448$

Answer and solution

Answer: (C) $224$

$t_n = a r^{n-1} = 7 \cdot 2^{5} = 224$
The sum of the first $4$ terms of $2 + 10 + 50 + \dots$ is
A $312$ B $1248$ C $1562$ D $1250$

Answer and solution

Answer: (A) $312$

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 2 \cdot \frac{625 - 1}{4} = 312$
The sum to infinity of $\displaystyle 1 + 2 \cdot \frac{1}{5} + 3 \cdot \frac{1}{25} + 4 \cdot \frac{1}{125} + \dots$ is
A $\displaystyle \frac{3}{2}$ B $\displaystyle \frac{5}{4}$ C $\displaystyle \frac{45}{16}$ D $\displaystyle \frac{25}{16}$

Answer and solution

Answer: (D) $\displaystyle \frac{25}{16}$

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

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{5}{4} + \frac{5}{16} = \frac{25}{16}$
The $n$ th term of $6, 14, 24, 36, \dots$ is
A $n^{2} + 7 n + 6$ B $n^{2} + 6 n - 2$ C $n^{2} + 5 n$ D $2 n^{2} + 4 n - 1$

Answer and solution

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

Differences: $8, 10, 12$ .

Fitting $n = 1, 2, 3$ gives $T_n = n^{2} + 5 n$ .
The $n$ th term of $11, 21, 41, 81, \dots$ is
A $5 \cdot 2^{n}$ B $5 \cdot 2^{n + 1} + 1$ C $5 \cdot 2^{n} - 1$ D $5 \cdot 2^{n} + 1$

Answer and solution

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

Differences: $10, 20, 40$ .

Fitting $n = 1, 2, 3$ gives $T_n = 5 \cdot 2^{n} + 1$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $1^2 + 4^2 + 7^2 + \dots$ is
A $9 n^{2} - 11 n + 3$ B $(3 n - 2)^{2}$ C $(3 n - 5)^{2}$ D $(3 n + 1)^{2}$

Answer and solution

Answer: (B) $(3 n - 2)^{2}$

$T_1 = 1$ , $T_2 = 16$ .

Only $(3 n - 2)^{2}$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $5 + 11 + 17 + \dots$ is
A $n (3 n + 2)$ B $3 n^{2} - 4 n + 6$ C $3 n^{2} + 3 n - 1$ D $n (6 n - 1)$

Answer and solution

Answer: (A) $n (3 n + 2)$

$S_1 = 5$ , $S_2 = 16$ , $S_3 = 33$ .

Only $n (3 n + 2)$ gives all three.
The largest number that divides $3^{3 n} - 1$ for every $n \in \mathbb{N}$ is
A $364$ B $26$ C $52$ D $27$

Answer and solution

Answer: (B) $26$

$n = 1:\ 26,\ n = 2:\ 728$

$\text{G.C.D.}(26, 728) = 26$
The largest number that divides $7^{2 n} - 1$ for every $n \in \mathbb{N}$ is
A $49$ B $1200$ C $48$ D $96$

Answer and solution

Answer: (C) $48$

$n = 1:\ 48,\ n = 2:\ 2400$

$\text{G.C.D.}(48, 2400) = 48$
To prove $10^n - 7^n$ is divisible by $3$ , assume $10^k - 7^k = 3\lambda$ . Then $10^{k+1} - 7^{k+1} =$
A $3(\lambda + 7^k)$ B $3(10\lambda + 7^k)$ C $3(7\lambda + 10^k)$ D $3(10\lambda - 7^k)$

Answer and solution

Answer: (B) $3(10\lambda + 7^k)$

$10^{k+1} - 7^{k+1} = 10(3\lambda + 7^k) - 7 \cdot 7^k$

$= 30\lambda + 3 \cdot 7^k = 3(10\lambda + 7^k)$
For all positive integers $n$ , $7^n - 4^n$ is divisible by
A $3$ B $11$ C $28$ D $4$

Answer and solution

Answer: (A) $3$

$x^n - y^n$ is divisible by $x - y$ .

$7 - 4 = 3$ .
If $150 \equiv r \pmod{17}$ with $0 \le r < 17$ , then $r =$
A $8$ B $3$ C $15$ D $14$

Answer and solution

Answer: (D) $14$

$150 = 17 \times 8 + 14$

$\therefore 150 - 14 = 136$ is divisible by $17$ , so $r = 14$ .
$5^{2n-1} + 2^{2n-1}$ is divisible by
A $3$ B $10$ C $7$ D $29$

Answer and solution

Answer: (C) $7$

$x^{2n-1} + y^{2n-1}$ is divisible by $x + y = 5 + 2 = 7$ .

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MCQ test, Set 6