If $P(1)$ is true and $P(k)$ true implies $P(k+1)$ true for every $k \in \mathbb{N}$, then

$P(n)$ is true for all $n \in \mathbb{N}$

$P(n)$ is true only for $n = 1$

$P(n)$ is true only for $n = k$

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

$\frac{n (2 n^{2} + 9 n + 13)}{6}$

$\frac{(n - 1) (2 n^{2} + 5 n + 6)}{6}$

$\frac{n (2 n^{2} + 9 n + 13)}{3}$

$\frac{(n + 1) (2 n^{2} + 13 n + 24)}{6}$

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

$\frac{n (4 n^{2} + 21 n + 29)}{6}$

$\frac{n (4 n^{2} + 21 n + 29)}{3}$

$\frac{(n + 1) (4 n^{2} + 29 n + 54)}{6}$

$\frac{(n - 1) (4 n^{2} + 13 n + 12)}{6}$

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

$\frac{n (n^{2} + 6 n + 11)}{3}$

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

$\frac{n^{3} + 6 n^{2} + 14 n - 3}{3}$

Maths 1A: Mathematical Induction: MCQ test, Set 4

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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 $5 + 7 + 9 + \dots + (2 n + 3) = n (n + 4)$ . In $P(3)$ , the left side $5 + 7 + 9$ equals
A $32$ B $21$ C $12$ D $9$

Answer and solution

Answer: (B) $21$

$5 + 7 + 9 = 21$

and the right side at $n = 3$ is also $21$ .
If $P(1)$ is true and $P(k)$ true implies $P(k+1)$ true for every $k \in \mathbb{N}$ , then
A $P(n)$ is false for some $n$ B $P(n)$ is true only for $n = 1$ C $P(n)$ is true only for $n = k$ D $P(n)$ is true for all $n \in \mathbb{N}$

Answer and solution

Answer: (D) $P(n)$ is true for all $n \in \mathbb{N}$

These are the two conditions of the principle of finite induction.

So $P(n)$ holds for every natural number.
To prove $\displaystyle 6 + 36 + 216 + \dots + 6 \cdot 6^{n - 1} = \frac{6 (6^{n} - 1)}{5}$ by induction, the term added to both sides of $P(k)$ is
A $6 \cdot 6^{k}$ B $6 \cdot 6^{k + 1}$ C $6 \cdot 6^{k - 1}$ D $6 \cdot 6^{k - 1} + 1$

Answer and solution

Answer: (A) $6 \cdot 6^{k}$

The $(k+1)$ th term is got by putting $n = k + 1$ in $6 \cdot 6^{n - 1}$ .

It is $6 \cdot 6^{k}$ .
By induction it can be shown that $2^2 + 3^2 + 4^2 + \dots + (n + 1)^{2} =$
A $\displaystyle \frac{(n - 1) (2 n^{2} + 5 n + 6)}{6}$ B $\displaystyle \frac{n (2 n^{2} + 9 n + 13)}{3}$ C $\displaystyle \frac{(n + 1) (2 n^{2} + 13 n + 24)}{6}$ D $\displaystyle \frac{n (2 n^{2} + 9 n + 13)}{6}$

Answer and solution

Answer: (D) $\displaystyle \frac{n (2 n^{2} + 9 n + 13)}{6}$

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

The formula $\displaystyle \frac{n (2 n^{2} + 9 n + 13)}{6}$ is the one proved by induction.
$1 + 2 + 3 + \dots + 55 =$
A $1540$ B $1596$ C $3080$ D $1485$

Answer and solution

Answer: (A) $1540$

$\displaystyle \frac{n(n+1)}{2} = \frac{55 \cdot 56}{2} = 1540$
$\displaystyle \sum_{k=1}^{9} k^3 =$
A $1296$ B $4050$ C $285$ D $2025$

Answer and solution

Answer: (D) $2025$

$\displaystyle \left(\frac{n(n+1)}{2}\right)^2 = 45^2 = 2025$
The sum of the first $37$ odd natural numbers is
A $1369$ B $1406$ C $1444$ D $1368$

Answer and solution

Answer: (A) $1369$

$1 + 3 + \dots + (2n - 1) = n^2 = 1369$
The sum of the first $45$ even natural numbers is
A $2070$ B $4050$ C $1980$ D $2025$

Answer and solution

Answer: (A) $2070$

$2 + 4 + \dots + 2n = n(n+1) = 45 \cdot 46 = 2070$
$1^2 - 2^2 + 3^2 - \dots + 13^2 =$
A $-78$ B $182$ C $91$ D $-91$

Answer and solution

Answer: (C) $91$

$n = 13$ is odd.

$\displaystyle \frac{13 \cdot 14}{2} = 91$
$\displaystyle \sum_{k=1}^{n} (k + 2) (2 k + 1) =$
A $\displaystyle \frac{n (4 n^{2} + 21 n + 29)}{3}$ B $\displaystyle \frac{(n + 1) (4 n^{2} + 29 n + 54)}{6}$ C $\displaystyle \frac{(n - 1) (4 n^{2} + 13 n + 12)}{6}$ D $\displaystyle \frac{n (4 n^{2} + 21 n + 29)}{6}$

Answer and solution

Answer: (D) $\displaystyle \frac{n (4 n^{2} + 21 n + 29)}{6}$

Expand $(k + 2) (2 k + 1) = 2 k^{2} + 5 k + 2$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (4 n^{2} + 21 n + 29)}{6}$ .
$\displaystyle \sum_{k=1}^{6} k^{2} + 3 k =$
A $224$ B $324$ C $154$ D $100$

Answer and solution

Answer: (C) $154$

Expand $k^{2} + 3 k = k^{2} + 3 k$ and use the standard sums with $n = 6$ .

The sum is $154$ .
The sum of the first $10$ terms of the A.P. $2, 11, 20, \dots$ is
A $470$ B $415$ C $830$ D $425$

Answer and solution

Answer: (D) $425$

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{10}{2}[4 + 81] = 425$
The sum to infinity of $\displaystyle 18 + 6 + 2 + \frac{2}{3} + \dots$ is
A $12$ B $27$ C $54$ D $\displaystyle \frac{27}{2}$

Answer and solution

Answer: (B) $27$

$a = 18$ , $\displaystyle r = \frac{1}{3}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = 27$
The $5$ th term of the G.P. $3, 15, 75, \dots$ is
A $50625$ B $9375$ C $1875$ D $375$

Answer and solution

Answer: (C) $1875$

$t_n = a r^{n-1} = 3 \cdot 5^{4} = 1875$
The sum to infinity of $\displaystyle 2 + 4 \cdot \frac{1}{3} + 6 \cdot \frac{1}{9} + 8 \cdot \frac{1}{27} + \dots$ is
A $\displaystyle \frac{9}{2}$ B $3$ C $\displaystyle \frac{15}{2}$ D $4$

Answer and solution

Answer: (A) $\displaystyle \frac{9}{2}$

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

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = 3 + \frac{3}{2} = \frac{9}{2}$
The $n$ th term of $2, 11, 26, 47, \dots$ is
A $3 n^{2} + 6 n + 2$ B $3 n^{2}$ C $3 n^{2} - 1$ D $4 n^{2} - n - 2$

Answer and solution

Answer: (C) $3 n^{2} - 1$

Differences: $9, 15, 21$ .

Fitting $n = 1, 2, 3$ gives $T_n = 3 n^{2} - 1$ .
The $n$ th term of $10, 22, 58, 166, \dots$ is
A $2 \cdot 3^{n} - 4$ B $2 \cdot 3^{n} + 4$ C $2 \cdot 3^{n}$ D $2 \cdot 3^{n + 1} + 4$

Answer and solution

Answer: (B) $2 \cdot 3^{n} + 4$

Differences: $12, 36, 108$ .

Fitting $n = 1, 2, 3$ gives $T_n = 2 \cdot 3^{n} + 4$ .
Using $n = 1$ and $n = 2$ , the $n$ th term of $4 + 10 + 16 + \dots$ is
A $6 n - 2$ B $7 n - 3$ C $6 n + 4$ D $6 n - 8$

Answer and solution

Answer: (A) $6 n - 2$

$T_1 = 4$ , $T_2 = 10$ .

Only $6 n - 2$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \dots$ is
A $n (n + 1) (n + 2)$ B $\displaystyle \frac{n^{3} + 3 n^{2} + 2 n + 12}{3}$ C $\displaystyle \frac{n (n^{2} + 6 n + 11)}{3}$ D $\displaystyle \frac{n^{3} + 6 n^{2} + 14 n - 3}{3}$

Answer and solution

Answer: (C) $\displaystyle \frac{n (n^{2} + 6 n + 11)}{3}$

$S_1 = 6$ , $S_2 = 18$ , $S_3 = 38$ .

Only $\displaystyle \frac{n (n^{2} + 6 n + 11)}{3}$ gives all three.
The largest number that divides $15^{n} - 8^{n}$ for every $n \in \mathbb{N}$ is
A $21$ B $7$ C $14$ D $8$

Answer and solution

Answer: (B) $7$

$n = 1:\ 7,\ n = 2:\ 161$

$\text{G.C.D.}(7, 161) = 7$
The largest number that divides $5^{n} - 1^{n}$ for every $n \in \mathbb{N}$ is
A $12$ B $4$ C $5$ D $8$

Answer and solution

Answer: (B) $4$

$n = 1:\ 4,\ n = 2:\ 24$

$\text{G.C.D.}(4, 24) = 4$
To prove $8^n - 3^n$ is divisible by $5$ , assume $8^k - 3^k = 5\lambda$ . Then $8^{k+1} - 3^{k+1} =$
A $5(8\lambda - 3^k)$ B $5(3\lambda + 8^k)$ C $5(\lambda + 3^k)$ D $5(8\lambda + 3^k)$

Answer and solution

Answer: (D) $5(8\lambda + 3^k)$

$8^{k+1} - 3^{k+1} = 8(5\lambda + 3^k) - 3 \cdot 3^k$

$= 40\lambda + 5 \cdot 3^k = 5(8\lambda + 3^k)$
With $P = 3$ , $P^{n+1} + (P + 1)^{2n-1}$ is divisible by
A $12$ B $13$ C $16$ D $7$

Answer and solution

Answer: (B) $13$

$P^{n+1} + (P+1)^{2n-1}$ is divisible by $P^2 + P + 1 = 9 + 3 + 1 = 13$ .
If $123 \equiv r \pmod{10}$ with $0 \le r < 10$ , then $r =$
A $3$ B $7$ C $4$ D $12$

Answer and solution

Answer: (A) $3$

$123 = 10 \times 12 + 3$

$\therefore 123 - 3 = 120$ is divisible by $10$ , so $r = 3$ .
By Fermat's result, $n^{11} - n$ is divisible by
A $22$ B $12$ C $11$ D $10$

Answer and solution

Answer: (C) $11$

For a prime $p$ , $n^p - n$ is divisible by $p$ ; here $p = 11$ .

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MCQ test, Set 4