The last step of a proof by induction is

concluding that $P(n)$ holds for all $n \in \mathbb{N}$

By induction it can be shown that $3^2 + 7^2 + 11^2 + \dots + (4 n - 1)^{2} =$

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

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

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

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

$\sum_{k=1}^{n} (k + 2) (k + 4) =$

$\frac{n (2 n^{2} + 21 n + 67)}{6}$

$\frac{(n + 1) (2 n^{2} + 25 n + 90)}{6}$

$\frac{n (2 n^{2} + 21 n + 67)}{3}$

$\frac{(n - 1) (2 n^{2} + 17 n + 48)}{6}$

Testing with $n = 1, 2, 3$, the sum of $n$ terms of $1 \cdot 4 + 2 \cdot 5 + 3 \cdot 6 + \dots$ is

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

$\frac{n^{3} + 3 n^{2} - 4 n + 12}{3}$

Maths 1A: Mathematical Induction: MCQ test, Set 10

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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 $5 + 13 + 21 + \dots + (8 n - 3) = n (4 n + 1)$ . In $P(3)$ , the left side $5 + 13 + 21$ equals
A $68$ B $21$ C $18$ D $39$

Answer and solution

Answer: (D) $39$

$5 + 13 + 21 = 39$

and the right side at $n = 3$ is also $39$ .
The last step of a proof by induction is
A adding the $(k+1)$ th term B concluding that $P(n)$ holds for all $n \in \mathbb{N}$ C assuming $P(k)$ D verifying $P(1)$

Answer and solution

Answer: (B) concluding that $P(n)$ holds for all $n \in \mathbb{N}$

After the base and inductive steps, we conclude by the principle of induction.
To prove $\displaystyle 8 + 13 + 18 + \dots + (5 n + 3) = \frac{n (5 n + 11)}{2}$ by induction, the term added to both sides of $P(k)$ is
A $5 k + 3$ B $5 k + 13$ C $5 k + 8$ D $5 k + 4$

Answer and solution

Answer: (C) $5 k + 8$

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

It is $5 k + 8$ .
By induction it can be shown that $3^2 + 7^2 + 11^2 + \dots + (4 n - 1)^{2} =$
A $\displaystyle \frac{n (16 n^{2} + 12 n - 1)}{3}$ B $\displaystyle \frac{2 n (16 n^{2} + 12 n - 1)}{3}$ C $\displaystyle \frac{(n - 1) (16 n^{2} - 20 n + 3)}{3}$ D $\displaystyle \frac{(n + 1) (16 n^{2} + 44 n + 27)}{3}$

Answer and solution

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

Check with $n = 1$ : the sum is $9$ and the formula gives $9$ .

The formula $\displaystyle \frac{n (16 n^{2} + 12 n - 1)}{3}$ is the one proved by induction.
$\displaystyle \sum_{k=1}^{11} k^3 =$
A $3025$ B $4356$ C $506$ D $2178$

Answer and solution

Answer: (B) $4356$

$\displaystyle \left(\frac{n(n+1)}{2}\right)^2 = 66^2 = 4356$
$1^2 + 2^2 + \dots + 9^2 =$
A $285$ B $255$ C $570$ D $2025$

Answer and solution

Answer: (A) $285$

$\displaystyle \frac{n(n+1)(2n+1)}{6} = \frac{9 \cdot 10 \cdot 19}{6} = 285$
$1^2 + 3^2 + 5^2 + \dots$ to $14$ terms $=$
A $6930$ B $3682$ C $4060$ D $3654$

Answer and solution

Answer: (D) $3654$

$\displaystyle \frac{n}{3}(4n^2 - 1) = \frac{14}{3}(783) = 3654$
$2^2 + 4^2 + \dots + 28^2 =$
A $4060$ B $1015$ C $2030$ D $7714$

Answer and solution

Answer: (A) $4060$

$\displaystyle 4\sum_{k=1}^{14} k^2 = 4 \cdot 1015 = 4060$
$1^2 - 2^2 + 3^2 - \dots - 8^2 =$
A $-72$ B $-36$ C $28$ D $36$

Answer and solution

Answer: (B) $-36$

$n = 8$ is even.

$\displaystyle -\frac{8 \cdot 9}{2} = -36$
$\displaystyle \sum_{k=1}^{n} (k + 2) (k + 4) =$
A $\displaystyle \frac{n (2 n^{2} + 21 n + 67)}{6}$ B $\displaystyle \frac{(n + 1) (2 n^{2} + 25 n + 90)}{6}$ C $\displaystyle \frac{n (2 n^{2} + 21 n + 67)}{3}$ D $\displaystyle \frac{(n - 1) (2 n^{2} + 17 n + 48)}{6}$

Answer and solution

Answer: (A) $\displaystyle \frac{n (2 n^{2} + 21 n + 67)}{6}$

Expand $(k + 2) (k + 4) = k^{2} + 6 k + 8$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (2 n^{2} + 21 n + 67)}{6}$ .
$\displaystyle \sum_{k=1}^{7} 4 k^{2} + 1 =$
A $824$ B $370$ C $567$ D $1379$

Answer and solution

Answer: (C) $567$

Expand $4 k^{2} + 1 = 4 k^{2} + 1$ and use the standard sums with $n = 7$ .

The sum is $567$ .
The sum of the first $11$ terms of the A.P. $7, 11, 15, \dots$ is
A $\displaystyle \frac{517}{2}$ B $517$ C $297$ D $319$

Answer and solution

Answer: (C) $297$

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{11}{2}[14 + 40] = 297$
The sum to infinity of $24 + 12 + 6 + 3 + \dots$ is
A $24$ B $12$ C $16$ D $48$

Answer and solution

Answer: (D) $48$

$a = 24$ , $\displaystyle r = \frac{1}{2}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = 48$
The $8$ th term of the G.P. $9, 18, 36, \dots$ is
A $576$ B $612220032$ C $2304$ D $1152$

Answer and solution

Answer: (D) $1152$

$t_n = a r^{n-1} = 9 \cdot 2^{7} = 1152$
The sum to infinity of $\displaystyle 1 + 7 \cdot \frac{1}{2} + 13 \cdot \frac{1}{4} + 19 \cdot \frac{1}{8} + \dots$ is
A $14$ B $8$ C $2$ D $26$

Answer and solution

Answer: (A) $14$

$a = 1$ , $d = 6$ , $\displaystyle r = \frac{1}{2}$ .

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = 2 + 12 = 14$
The $n$ th term of $6, 12, 20, 30, \dots$ is
A $n^{2} + 5 n + 6$ B $n^{2} + 3 n$ C $2 n^{2} + 2 n + 1$ D $n^{2} + 3 n + 2$

Answer and solution

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

Differences: $6, 8, 10$ .

Fitting $n = 1, 2, 3$ gives $T_n = n^{2} + 3 n + 2$ .
The $n$ th term of $9, 51, 345, 2403, \dots$ is
A $7^{n}$ B $7^{n} + 2$ C $7^{n + 1} + 2$ D $7^{n} - 2$

Answer and solution

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

Differences: $42, 294, 2058$ .

Fitting $n = 1, 2, 3$ gives $T_n = 7^{n} + 2$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $3^2 + 7^2 + 11^2 + \dots$ is
A $(4 n + 3)^{2}$ B $(4 n - 5)^{2}$ C $(4 n - 1)^{2}$ D $16 n^{2} - 7 n$

Answer and solution

Answer: (C) $(4 n - 1)^{2}$

$T_1 = 9$ , $T_2 = 49$ .

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

Answer and solution

Answer: (B) $\displaystyle \frac{n (n + 1) (n + 5)}{3}$

$S_1 = 4$ , $S_2 = 14$ , $S_3 = 32$ .

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

Answer and solution

Answer: (A) $15$

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

$\text{G.C.D.}(15, 255) = 15$
The largest number that divides $16^{n} - 9^{n}$ for every $n \in \mathbb{N}$ is
A $8$ B $14$ C $7$ D $21$

Answer and solution

Answer: (C) $7$

$n = 1:\ 7,\ n = 2:\ 175$

$\text{G.C.D.}(7, 175) = 7$
To prove $17^n - 3^n$ is divisible by $14$ , assume $17^k - 3^k = 14\lambda$ . Then $17^{k+1} - 3^{k+1} =$
A $14(17\lambda - 3^k)$ B $14(3\lambda + 17^k)$ C $14(\lambda + 3^k)$ D $14(17\lambda + 3^k)$

Answer and solution

Answer: (D) $14(17\lambda + 3^k)$

$17^{k+1} - 3^{k+1} = 17(14\lambda + 3^k) - 3 \cdot 3^k$

$= 238\lambda + 14 \cdot 3^k = 14(17\lambda + 3^k)$
If $58 \equiv r \pmod{11}$ with $0 \le r < 11$ , then $r =$
A $3$ B $5$ C $8$ D $4$

Answer and solution

Answer: (A) $3$

$58 = 11 \times 5 + 3$

$\therefore 58 - 3 = 55$ is divisible by $11$ , so $r = 3$ .
If $127 \equiv r \pmod{9}$ with $0 \le r < 9$ , then $r =$
A $14$ B $1$ C $2$ D $8$

Answer and solution

Answer: (B) $1$

$127 = 9 \times 14 + 1$

$\therefore 127 - 1 = 126$ is divisible by $9$ , so $r = 1$ .
If $91 \equiv r \pmod{16}$ with $0 \le r < 16$ , then $r =$
A $5$ B $13$ C $12$ D $11$

Answer and solution

Answer: (D) $11$

$91 = 16 \times 5 + 11$

$\therefore 91 - 11 = 80$ is divisible by $16$ , so $r = 11$ .

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MCQ test, Set 10