By induction it can be shown that $1^2 + 5^2 + 9^2 + \dots + (4 n - 3)^{2} =$

$\frac{n (16 n^{2} - 12 n - 1)}{3}$

$\frac{(n - 1) (16 n^{2} - 44 n + 27)}{3}$

$\frac{(n + 1) (16 n^{2} + 20 n + 3)}{3}$

$\frac{2 n (16 n^{2} - 12 n - 1)}{3}$

$\sum_{k=1}^{n} k (k + 6) =$

$\frac{n (n + 1) (2 n + 19)}{6}$

$\frac{n (n + 1) (2 n + 19)}{3}$

$\frac{(n + 1) (n + 2) (2 n + 21)}{6}$

$\frac{n (n - 1) (2 n + 17)}{6}$

Testing with $n = 1, 2, 3$, the sum of $n$ terms of $1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots$ is

$\frac{n (n + 1) (2 n + 7)}{6}$

$\frac{2 n^{3} + 3 n^{2} - 5 n + 18}{6}$

$\frac{2 n^{3} + 9 n^{2} + 13 n - 6}{6}$

Maths 1A: Mathematical Induction: MCQ test, Set 7

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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 $P(n)$ : $7 + 13 + 19 + \dots + (6 n + 1) = n (3 n + 4)$ , then the value of each side of $P(3)$ is
A $19$ B $20$ C $64$ D $39$

Answer and solution

Answer: (D) $39$

$7 + 13 + 19 = 39$

and the right side at $n = 3$ is also $39$ .
In this chapter $n$ takes values in $\mathbb{N}$ , which is
A $\{1, 2, 3, \dots\}$ B $\{2, 3, 4, \dots\}$ C $\{\dots, -1, 0, 1, \dots\}$ D $\{0, 1, 2, \dots\}$

Answer and solution

Answer: (A) $\{1, 2, 3, \dots\}$

$\mathbb{N} = \{1, 2, 3, \dots\}$ , the natural numbers.
To prove $\displaystyle 1^2 + 4^2 + 7^2 + \dots + (3 n - 2)^{2} = \frac{n (6 n^{2} - 3 n - 1)}{2}$ by induction, the term added to both sides of $P(k)$ is
A $(3 k + 4)^{2}$ B $(3 k - 2)^{2} + 1$ C $(3 k - 2)^{2}$ D $(3 k + 1)^{2}$

Answer and solution

Answer: (D) $(3 k + 1)^{2}$

The $(k+1)$ th term is got by putting $n = k + 1$ in $(3 n - 2)^{2}$ .

It is $(3 k + 1)^{2}$ .
By induction it can be shown that $1^2 + 5^2 + 9^2 + \dots + (4 n - 3)^{2} =$
A $\displaystyle \frac{(n - 1) (16 n^{2} - 44 n + 27)}{3}$ B $\displaystyle \frac{(n + 1) (16 n^{2} + 20 n + 3)}{3}$ C $\displaystyle \frac{n (16 n^{2} - 12 n - 1)}{3}$ D $\displaystyle \frac{2 n (16 n^{2} - 12 n - 1)}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{n (16 n^{2} - 12 n - 1)}{3}$

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

The formula $\displaystyle \frac{n (16 n^{2} - 12 n - 1)}{3}$ is the one proved by induction.
$1 + 2 + 3 + \dots + 85 =$
A $3570$ B $3655$ C $7310$ D $3741$

Answer and solution

Answer: (B) $3655$

$\displaystyle \frac{n(n+1)}{2} = \frac{85 \cdot 86}{2} = 3655$
$\displaystyle \sum_{k=1}^{21} k^3 =$
A $3311$ B $44100$ C $106722$ D $53361$

Answer and solution

Answer: (D) $53361$

$\displaystyle \left(\frac{n(n+1)}{2}\right)^2 = 231^2 = 53361$
The sum of the first $38$ even natural numbers is
A $1406$ B $2888$ C $1482$ D $1444$

Answer and solution

Answer: (C) $1482$

$2 + 4 + \dots + 2n = n(n+1) = 38 \cdot 39 = 1482$
$1^2 + 3^2 + 5^2 + \dots$ to $6$ terms $=$
A $506$ B $286$ C $364$ D $290$

Answer and solution

Answer: (B) $286$

$\displaystyle \frac{n}{3}(4n^2 - 1) = \frac{6}{3}(143) = 286$
$1^2 - 2^2 + 3^2 - \dots$ to $(2n + 1)$ terms $=$
A $n(2n+1)$ B $-(n+1)(2n+1)$ C $\displaystyle \frac{(n+1)(2n+1)}{2}$ D $(n+1)(2n+1)$

Answer and solution

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

$2n + 1$ is odd, so the sum is $\displaystyle \frac{(2n+1)(2n+2)}{2} = (n+1)(2n+1)$ .
$\displaystyle \sum_{k=1}^{n} k (k + 6) =$
A $\displaystyle \frac{n (n + 1) (2 n + 19)}{3}$ B $\displaystyle \frac{n (n + 1) (2 n + 19)}{6}$ C $\displaystyle \frac{(n + 1) (n + 2) (2 n + 21)}{6}$ D $\displaystyle \frac{n (n - 1) (2 n + 17)}{6}$

Answer and solution

Answer: (B) $\displaystyle \frac{n (n + 1) (2 n + 19)}{6}$

Expand $k (k + 6) = k^{2} + 6 k$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (n + 1) (2 n + 19)}{6}$ .
$\displaystyle \sum_{k=1}^{6} k^{3} + k =$
A $1332$ B $240$ C $812$ D $462$

Answer and solution

Answer: (D) $462$

Expand $k^{3} + k = k^{3} + k$ and use the standard sums with $n = 6$ .

The sum is $462$ .
The $9$ th term of the A.P. $4, 15, 26, \dots$ is
A $99$ B $103$ C $81$ D $92$

Answer and solution

Answer: (D) $92$

$t_n = a + (n-1)d = 4 + 8(11) = 92$
The sum to infinity of $\displaystyle 36 + 6 + 1 + \frac{1}{6} + \dots$ is
A $216$ B $30$ C $\displaystyle \frac{216}{5}$ D $\displaystyle \frac{216}{7}$

Answer and solution

Answer: (C) $\displaystyle \frac{216}{5}$

$a = 36$ , $\displaystyle r = \frac{1}{6}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = \frac{216}{5}$
The $5$ th term of the G.P. $6, 18, 54, \dots$ is
A $486$ B $104976$ C $162$ D $1458$

Answer and solution

Answer: (A) $486$

$t_n = a r^{n-1} = 6 \cdot 3^{4} = 486$
The sum to infinity of $\displaystyle 2 + 5 \cdot \frac{1}{2} + 8 \cdot \frac{1}{4} + 11 \cdot \frac{1}{8} + \dots$ is
A $10$ B $16$ C $4$ D $7$

Answer and solution

Answer: (A) $10$

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

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = 4 + 6 = 10$
The $n$ th term of $6, 17, 34, 57, \dots$ is
A $3 n^{2} + 8 n + 6$ B $3 n^{2} + 2 n + 1$ C $4 n^{2} + n$ D $3 n^{2} + 2 n$

Answer and solution

Answer: (B) $3 n^{2} + 2 n + 1$

Differences: $11, 17, 23$ .

Fitting $n = 1, 2, 3$ gives $T_n = 3 n^{2} + 2 n + 1$ .
The $n$ th term of $5, 35, 215, 1295, \dots$ is
A $6^{n}$ B $6^{n + 1} - 1$ C $6^{n} + 1$ D $6^{n} - 1$

Answer and solution

Answer: (D) $6^{n} - 1$

Differences: $30, 180, 1080$ .

Fitting $n = 1, 2, 3$ gives $T_n = 6^{n} - 1$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + 4 \cdot 5 \cdot 6 + \dots$ is
A $n^{3} + 6 n^{2} + 12 n + 5$ B $(n + 1) (n + 2) (n + 3)$ C $(n + 2) (n + 3) (n + 4)$ D $n (n + 1) (n + 2)$

Answer and solution

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

$T_1 = 24$ , $T_2 = 60$ .

Only $(n + 1) (n + 2) (n + 3)$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \dots$ is
A $n^{2} (n + 2)$ B $\displaystyle \frac{2 n^{3} + 3 n^{2} - 5 n + 18}{6}$ C $\displaystyle \frac{n (n + 1) (2 n + 7)}{6}$ D $\displaystyle \frac{2 n^{3} + 9 n^{2} + 13 n - 6}{6}$

Answer and solution

Answer: (C) $\displaystyle \frac{n (n + 1) (2 n + 7)}{6}$

$S_1 = 3$ , $S_2 = 11$ , $S_3 = 26$ .

Only $\displaystyle \frac{n (n + 1) (2 n + 7)}{6}$ gives all three.
The largest number that divides $11^{n} - 1$ for every $n \in \mathbb{N}$ is
A $10$ B $60$ C $11$ D $20$

Answer and solution

Answer: (A) $10$

$n = 1:\ 10,\ n = 2:\ 120$

$\text{G.C.D.}(10, 120) = 10$
The largest number that divides $17^{n} - 5^{n}$ for every $n \in \mathbb{N}$ is
A $13$ B $24$ C $12$ D $132$

Answer and solution

Answer: (C) $12$

$n = 1:\ 12,\ n = 2:\ 264$

$\text{G.C.D.}(12, 264) = 12$
To prove $11^n - 3^n$ is divisible by $8$ , assume $11^k - 3^k = 8\lambda$ . Then $11^{k+1} - 3^{k+1} =$
A $8(\lambda + 3^k)$ B $8(11\lambda + 3^k)$ C $8(3\lambda + 11^k)$ D $8(11\lambda - 3^k)$

Answer and solution

Answer: (B) $8(11\lambda + 3^k)$

$11^{k+1} - 3^{k+1} = 11(8\lambda + 3^k) - 3 \cdot 3^k$

$= 88\lambda + 8 \cdot 3^k = 8(11\lambda + 3^k)$
If $99 \equiv r \pmod{12}$ with $0 \le r < 12$ , then $r =$
A $8$ B $4$ C $3$ D $9$

Answer and solution

Answer: (C) $3$

$99 = 12 \times 8 + 3$

$\therefore 99 - 3 = 96$ is divisible by $12$ , so $r = 3$ .
The statement $8^n - 1$ is divisible by $7$ can be written as
A $8^n \equiv 0 \pmod{7}$ B $7^n \equiv 1 \pmod{8}$ C $8^n \equiv 1 \pmod{7}$ D $8^n \equiv 7 \pmod{1}$

Answer and solution

Answer: (C) $8^n \equiv 1 \pmod{7}$

$8^n - 1$ divisible by $7$ means $8^n \equiv 1 \pmod 7$ .
If $46 \equiv r \pmod{5}$ with $0 \le r < 5$ , then $r =$
A $1$ B $4$ C $9$ D $2$

Answer and solution

Answer: (A) $1$

$46 = 5 \times 9 + 1$

$\therefore 46 - 1 = 45$ is divisible by $5$ , so $r = 1$ .

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MCQ test, Set 7