In an induction proof, the assumption that $P(k)$ is true is called

the principle of complete induction

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

$\frac{2 n (n + 4) (n + 5)}{3}$

$\frac{(n + 1) (n + 5) (n + 6)}{3}$

$\frac{(n - 1) (n + 3) (n + 4)}{3}$

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

$\frac{n (n + 1) (4 n - 1)}{6}$

$\frac{4 n^{3} + 3 n^{2} + 5 n - 6}{6}$

$\frac{4 n^{3} - 9 n^{2} + 5 n + 6}{6}$

Maths 1A: Mathematical Induction: MCQ test, Set 2

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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 $\displaystyle 5 + 12 + 19 + \dots + (7 n - 2) = \frac{n (7 n + 3)}{2}$ . In $P(2)$ , the left side $5 + 12$ equals
A $5$ B $36$ C $17$ D $12$

Answer and solution

Answer: (C) $17$

$5 + 12 = 17$

and the right side at $n = 2$ is also $17$ .
In an induction proof, the assumption that $P(k)$ is true is called
A the induction hypothesis B the principle of complete induction C the base step D the conclusion

Answer and solution

Answer: (A) the induction hypothesis

Inside the inductive step we assume $P(k)$ ; this assumption is the induction hypothesis.
To prove $7 + 11 + 15 + \dots + (4 n + 3) = n (2 n + 5)$ by induction, the term added to both sides of $P(k)$ is
A $4 k + 3$ B $4 k + 7$ C $4 k + 11$ D $4 k + 4$

Answer and solution

Answer: (B) $4 k + 7$

The $(k+1)$ th term is got by putting $n = k + 1$ in $(4 n + 3)$ .

It is $4 k + 7$ .
By induction it can be shown that $\displaystyle \frac{1}{3 \cdot 7} + \frac{1}{7 \cdot 11} + \frac{1}{11 \cdot 15} + \dots + \frac{1}{(4 n - 1)(4 n + 3)} =$
A $\displaystyle \frac{n}{12 n + 9}$ B $\displaystyle \frac{n + 1}{3 (4 n + 7)}$ C $\displaystyle \frac{2 n}{3 (4 n + 3)}$ D $\displaystyle \frac{n - 1}{3 (4 n - 1)}$

Answer and solution

Answer: (A) $\displaystyle \frac{n}{12 n + 9}$

Check with $n = 1$ : the sum is $\displaystyle \frac{1}{21}$ and the formula gives $\displaystyle \frac{1}{21}$ .

The formula $\displaystyle \frac{n}{12 n + 9}$ is the one proved by induction.
$\displaystyle \sum_{k=1}^{13} k^3 =$
A $819$ B $16562$ C $6084$ D $8281$

Answer and solution

Answer: (D) $8281$

$\displaystyle \left(\frac{n(n+1)}{2}\right)^2 = 91^2 = 8281$
$1^4 + 2^4 + \dots + 4^4 =$
A $900$ B $354$ C $200$ D $98$

Answer and solution

Answer: (B) $354$

$\displaystyle \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$ with $n = 4$

$= 354$
$2^2 + 4^2 + \dots + 14^2 =$
A $1015$ B $280$ C $140$ D $560$

Answer and solution

Answer: (D) $560$

$\displaystyle 4\sum_{k=1}^{7} k^2 = 4 \cdot 140 = 560$
$1^3 + 3^3 + 5^3 + \dots$ to $n$ terms $=$
A $n^2(n+1)^2$ B $n^2(2n^2 + 1)$ C $\displaystyle \frac{n}{3}(4n^2 - 1)$ D $n^2(2n^2 - 1)$

Answer and solution

Answer: (D) $n^2(2n^2 - 1)$

The standard result for the cubes of odd numbers.

$\displaystyle \sum (2k-1)^3 = n^2(2n^2 - 1)$
$1^2 - 2^2 + 3^2 - \dots - 20^2 =$
A $-210$ B $190$ C $-420$ D $210$

Answer and solution

Answer: (A) $-210$

$n = 20$ is even.

$\displaystyle -\frac{20 \cdot 21}{2} = -210$
$\displaystyle \sum_{k=1}^{n} (k + 1) (k + 4) =$
A $\displaystyle \frac{2 n (n + 4) (n + 5)}{3}$ B $\displaystyle \frac{n (n + 4) (n + 5)}{3}$ C $\displaystyle \frac{(n + 1) (n + 5) (n + 6)}{3}$ D $\displaystyle \frac{(n - 1) (n + 3) (n + 4)}{3}$

Answer and solution

Answer: (B) $\displaystyle \frac{n (n + 4) (n + 5)}{3}$

Expand $(k + 1) (k + 4) = k^{2} + 5 k + 4$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (n + 4) (n + 5)}{3}$ .
$\displaystyle \sum_{k=1}^{9} k (k - 1) =$
A $648$ 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 $n = 9$ .

The sum is $240$ .
The sum of the first $12$ terms of the A.P. $3, 8, 13, \dots$ is
A $396$ B $696$ C $348$ D $366$

Answer and solution

Answer: (D) $366$

$\displaystyle S_n = \frac{n}{2}[2a + (n-1)d] = \frac{12}{2}[6 + 55] = 366$
The sum of the first $5$ terms of $3 + 12 + 48 + \dots$ is
A $3069$ B $3072$ C $1023$ D $4095$

Answer and solution

Answer: (C) $1023$

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 3 \cdot \frac{1024 - 1}{3} = 1023$
The sum to infinity of $\displaystyle 20 - 5 + \frac{5}{4} - \frac{5}{16} + \dots$ is
A $\displaystyle \frac{80}{3}$ B $25$ C $16$ D $-80$

Answer and solution

Answer: (C) $16$

$a = 20$ , $\displaystyle r = - \frac{1}{4}$ .

$\displaystyle S_\infty = \frac{a}{1 - r} = 16$
The sum to infinity of $\displaystyle 3 + 4 \cdot \frac{1}{3} + 5 \cdot \frac{1}{9} + 6 \cdot \frac{1}{27} + \dots$ is
A $\displaystyle \frac{9}{2}$ B $\displaystyle \frac{21}{4}$ C $\displaystyle \frac{27}{4}$ D $5$

Answer and solution

Answer: (B) $\displaystyle \frac{21}{4}$

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

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{9}{2} + \frac{3}{4} = \frac{21}{4}$
The $n$ th term of $5, 11, 21, 35, \dots$ is
A $3 n^{2} - n + 2$ B $2 n^{2} + 4 n + 5$ C $2 n^{2} + 3$ D $2 n^{2}$

Answer and solution

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

Differences: $6, 10, 14$ .

Fitting $n = 1, 2, 3$ gives $T_n = 2 n^{2} + 3$ .
The $n$ th term of $10, 28, 82, 244, \dots$ is
A $3 \cdot 3^{n}$ B $3 \cdot 3^{n + 1} + 1$ C $3 \cdot 3^{n} - 1$ D $3 \cdot 3^{n} + 1$

Answer and solution

Answer: (D) $3 \cdot 3^{n} + 1$

Differences: $18, 54, 162$ .

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

Answer and solution

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

$T_1 = 9$ , $T_2 = 25$ .

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

Answer and solution

Answer: (B) $\displaystyle \frac{n (n + 1) (4 n - 1)}{6}$

$S_1 = 1$ , $S_2 = 7$ , $S_3 = 22$ .

Only $\displaystyle \frac{n (n + 1) (4 n - 1)}{6}$ gives all three.
The largest number that divides $12^{n} - 7^{n}$ for every $n \in \mathbb{N}$ is
A $15$ B $6$ C $10$ D $5$

Answer and solution

Answer: (D) $5$

$n = 1:\ 5,\ n = 2:\ 95$

$\text{G.C.D.}(5, 95) = 5$
The largest number that divides $4^{n} - 1$ for every $n \in \mathbb{N}$ is
A $4$ B $6$ C $3$ D $9$

Answer and solution

Answer: (C) $3$

$n = 1:\ 3,\ n = 2:\ 15$

$\text{G.C.D.}(3, 15) = 3$
To prove $9^n - 5^n$ is divisible by $4$ , assume $9^k - 5^k = 4\lambda$ . Then $9^{k+1} - 5^{k+1} =$
A $4(\lambda + 5^k)$ B $4(9\lambda + 5^k)$ C $4(5\lambda + 9^k)$ D $4(9\lambda - 5^k)$

Answer and solution

Answer: (B) $4(9\lambda + 5^k)$

$9^{k+1} - 5^{k+1} = 9(4\lambda + 5^k) - 5 \cdot 5^k$

$= 36\lambda + 4 \cdot 5^k = 4(9\lambda + 5^k)$
If $n$ is any odd integer, $n(n^2 - 1)$ is always divisible by
A $36$ B $16$ C $48$ D $24$

Answer and solution

Answer: (D) $24$

Standard result: for odd $n$ , $n(n^2 - 1)$ is divisible by $24$ .

For example $n = 5$ gives $120 = 24 \times 5$ .
If $100 \equiv r \pmod{7}$ with $0 \le r < 7$ , then $r =$
A $2$ B $3$ C $5$ D $14$

Answer and solution

Answer: (A) $2$

$100 = 7 \times 14 + 2$

$\therefore 100 - 2 = 98$ is divisible by $7$ , so $r = 2$ .
The product of $4$ consecutive natural numbers is always divisible by
A $24$ B $12$ C $16$ D $48$

Answer and solution

Answer: (A) $24$

The product of $n$ consecutive natural numbers is divisible by $n!$ ; here $4! = 24$ .

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MCQ test, Set 2