By induction it can be shown that $2^2 + 5^2 + 8^2 + \dots + (3 n - 1)^{2} =$

$\frac{n (6 n^{2} + 3 n - 1)}{2}$

$\frac{(n + 1) (6 n^{2} + 15 n + 8)}{2}$

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

$\sum_{k=1}^{n} k (k + 5) =$

$\frac{(n + 1) (n + 2) (n + 9)}{3}$

$\frac{2 n (n + 1) (n + 8)}{3}$

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

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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 $P(n)$ : $1 + 7 + 13 + \dots + (6 n - 5) = n (3 n - 2)$ , then the value of each side of $P(3)$ is
A $13$ B $8$ C $40$ D $21$

Answer and solution

Answer: (D) $21$

$1 + 7 + 13 = 21$

and the right side at $n = 3$ is also $21$ .
In the principle of finite mathematical induction, the base step is to check that
A $P(k+1)$ is true B $P(1)$ is true C $P(k)$ is true D $P(n)$ is true for every $n$

Answer and solution

Answer: (B) $P(1)$ is true

The base step verifies the first case, $P(1)$ , by direct calculation.
To prove $\displaystyle 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots + n(n + 2) = \frac{n (n + 1) (2 n + 7)}{6}$ by induction, the term added to both sides of $P(k)$ is
A $(k + 1) (k + 3)$ B $(k + 2) (k + 4)$ C $k (k + 2) + 1$ D $k (k + 2)$

Answer and solution

Answer: (A) $(k + 1) (k + 3)$

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

It is $(k + 1) (k + 3)$ .
By induction it can be shown that $2^2 + 5^2 + 8^2 + \dots + (3 n - 1)^{2} =$
A $\displaystyle \frac{(n + 1) (6 n^{2} + 15 n + 8)}{2}$ B $\displaystyle \frac{n (6 n^{2} + 3 n - 1)}{2}$ C $\displaystyle \frac{(n - 1) (6 n^{2} - 9 n + 2)}{2}$ D $n (6 n^{2} + 3 n - 1)$

Answer and solution

Answer: (B) $\displaystyle \frac{n (6 n^{2} + 3 n - 1)}{2}$

Check with $n = 1$ : the sum is $4$ and the formula gives $4$ .

The formula $\displaystyle \frac{n (6 n^{2} + 3 n - 1)}{2}$ is the one proved by induction.
$1 + 2 + 3 + \dots + 40 =$
A $780$ B $1640$ C $820$ D $861$

Answer and solution

Answer: (C) $820$

$\displaystyle \frac{n(n+1)}{2} = \frac{40 \cdot 41}{2} = 820$
$1^2 + 2^2 + \dots + 11^2 =$
A $1012$ B $4356$ C $506$ D $462$

Answer and solution

Answer: (C) $506$

$\displaystyle \frac{n(n+1)(2n+1)}{6} = \frac{11 \cdot 12 \cdot 23}{6} = 506$
$1^2 + 3^2 + 5^2 + \dots$ to $9$ terms $=$
A $1140$ B $1785$ C $975$ D $969$

Answer and solution

Answer: (D) $969$

$\displaystyle \frac{n}{3}(4n^2 - 1) = \frac{9}{3}(323) = 969$
$2^2 + 4^2 + 6^2 + \dots + (2n)^2 =$
A $\displaystyle \frac{2}{3}n(n+1)(2n+1)$ B $\displaystyle \frac{1}{6}n(n+1)(2n+1)$ C $\displaystyle \frac{n}{3}(4n^2 - 1)$ D $\displaystyle \frac{4}{3}n(n+1)(2n+1)$

Answer and solution

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

$(2k)^2 = 4k^2$

$\displaystyle 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2}{3}n(n+1)(2n+1)$
$1^2 - 2^2 + 3^2 - \dots + 11^2 =$
A $66$ B $132$ C $-66$ D $-55$

Answer and solution

Answer: (A) $66$

$n = 11$ is odd.

$\displaystyle \frac{11 \cdot 12}{2} = 66$
$\displaystyle \sum_{k=1}^{n} k (k + 5) =$
A $\displaystyle \frac{(n + 1) (n + 2) (n + 9)}{3}$ B $\displaystyle \frac{2 n (n + 1) (n + 8)}{3}$ C $\displaystyle \frac{n (n + 1) (n + 8)}{3}$ D $\displaystyle \frac{n (n - 1) (n + 7)}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{n (n + 1) (n + 8)}{3}$

Expand $k (k + 5) = k^{2} + 5 k$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (n + 1) (n + 8)}{3}$ .
$\displaystyle \sum_{k=1}^{8} k (k + 1) =$
A $576$ B $168$ C $330$ D $240$

Answer and solution

Answer: (D) $240$

Expand $k (k + 1) = k^{2} + k$ and use the standard sums with $n = 8$ .

The sum is $240$ .
The $20$ th term of the A.P. $6, 13, 20, \dots$ is
A $139$ B $140$ C $132$ D $146$

Answer and solution

Answer: (A) $139$

$t_n = a + (n-1)d = 6 + 19(7) = 139$
The $7$ th term of the G.P. $2, 6, 18, \dots$ is
A $4374$ B $1458$ C $46656$ D $486$

Answer and solution

Answer: (B) $1458$

$t_n = a r^{n-1} = 2 \cdot 3^{6} = 1458$
The sum to infinity of $\displaystyle 16 + 4 + 1 + \frac{1}{4} + \dots$ is
A $\displaystyle \frac{64}{5}$ B $64$ C $\displaystyle \frac{64}{3}$ D $12$

Answer and solution

Answer: (C) $\displaystyle \frac{64}{3}$

$a = 16$ , $\displaystyle r = \frac{1}{4}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = \frac{64}{3}$
The sum to infinity of $\displaystyle 1 + 3 \cdot \frac{1}{2} + 5 \cdot \frac{1}{4} + 7 \cdot \frac{1}{8} + \dots$ is
A $4$ B $2$ C $6$ D $10$

Answer and solution

Answer: (C) $6$

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

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = 2 + 4 = 6$
The $n$ th term of $3, 7, 13, 21, \dots$ is
A $n^{2} + n + 1$ B $n^{2} + n$ C $n^{2} + 3 n + 3$ D $2 n^{2}$

Answer and solution

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

Differences: $4, 6, 8$ .

Fitting $n = 1, 2, 3$ gives $T_n = n^{2} + n + 1$ .
The $n$ th term of $1, 3, 7, 15, \dots$ is
A $2^{n}$ B $2^{n} + 1$ C $2^{n} - 1$ D $2^{n + 1} - 1$

Answer and solution

Answer: (C) $2^{n} - 1$

Differences: $2, 4, 8$ .

Fitting $n = 1, 2, 3$ gives $T_n = 2^{n} - 1$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $1 \cdot 4 + 4 \cdot 7 + 7 \cdot 10 + \dots$ is
A $(3 n - 5) (3 n - 2)$ B $(3 n - 2) (3 n + 1)$ C $9 n^{2} - 2 n - 3$ D $(3 n + 1) (3 n + 4)$

Answer and solution

Answer: (B) $(3 n - 2) (3 n + 1)$

$T_1 = 4$ , $T_2 = 28$ .

Only $(3 n - 2) (3 n + 1)$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $1 + 5 + 9 + \dots$ is
A $2 n^{2} - 1$ B $n (4 n - 3)$ C $n (2 n - 1)$ D $2 n^{2} - 5 n + 4$

Answer and solution

Answer: (C) $n (2 n - 1)$

$S_1 = 1$ , $S_2 = 6$ , $S_3 = 15$ .

Only $n (2 n - 1)$ gives all three.
The largest number that divides $10^{n} - 4^{n}$ for every $n \in \mathbb{N}$ is
A $7$ B $42$ C $12$ D $6$

Answer and solution

Answer: (D) $6$

$n = 1:\ 6,\ n = 2:\ 84$

$\text{G.C.D.}(6, 84) = 6$
The largest number that divides $3^{2 n} - 1$ for every $n \in \mathbb{N}$ is
A $9$ B $8$ C $40$ D $16$

Answer and solution

Answer: (B) $8$

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

$\text{G.C.D.}(8, 80) = 8$
To prove $7^n - 2^n$ is divisible by $5$ , assume $7^k - 2^k = 5\lambda$ . Then $7^{k+1} - 2^{k+1} =$
A $5(2\lambda + 7^k)$ B $5(\lambda + 2^k)$ C $5(7\lambda - 2^k)$ D $5(7\lambda + 2^k)$

Answer and solution

Answer: (D) $5(7\lambda + 2^k)$

$7^{k+1} - 2^{k+1} = 7(5\lambda + 2^k) - 2 \cdot 2^k$

$= 35\lambda + 5 \cdot 2^k = 5(7\lambda + 2^k)$
If $53 \equiv r \pmod{6}$ with $0 \le r < 6$ , then $r =$
A $5$ B $0$ C $1$ D $8$

Answer and solution

Answer: (A) $5$

$53 = 6 \times 8 + 5$

$\therefore 53 - 5 = 48$ is divisible by $6$ , so $r = 5$ .
For every positive integer $n$ , $x^{2n+1} + y^{2n+1}$ is divisible by
A $x - y$ B $x^2 + y^2$ C $x + y$ D $xy$

Answer and solution

Answer: (C) $x + y$

An odd power sum always has the factor $x + y$ .
If $79 \equiv r \pmod{8}$ with $0 \le r < 8$ , then $r =$
A $0$ B $7$ C $1$ D $9$

Answer and solution

Answer: (B) $7$

$79 = 8 \times 9 + 7$

$\therefore 79 - 7 = 72$ is divisible by $8$ , so $r = 7$ .

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MCQ test, Set 1