In the inductive step for a sum formula, we

add the $(k+1)$th term to both sides of $P(k)$

multiply both sides of $P(k)$ by $k + 1$

add the $k$th term to both sides of $P(k)$

$\sum_{k=1}^{n} k (2 k - 3) =$

$\frac{n (n + 1) (4 n - 7)}{6}$

$\frac{n (n - 1) (4 n - 11)}{6}$

$\frac{n (n + 1) (4 n - 7)}{3}$

$\frac{(n + 1) (n + 2) (4 n - 3)}{6}$

Testing with $n = 1, 2, 3$, the sum of $n$ terms of $5 + 12 + 19 + \dots$ is

$\frac{7 n^{2} - 11 n + 14}{2}$

Maths 1A: Mathematical Induction: MCQ test, Set 9

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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 $P(n)$ : $\displaystyle 1 + 8 + 15 + \dots + (7 n - 6) = \frac{n (7 n - 5)}{2}$ , then the value of each side of $P(2)$ is
A $1$ B $24$ C $8$ D $9$

Answer and solution

Answer: (D) $9$

$1 + 8 = 9$

and the right side at $n = 2$ is also $9$ .
In the inductive step for a sum formula, we
A multiply both sides of $P(k)$ by $k + 1$ B subtract $P(1)$ from $P(k)$ C add the $(k+1)$ th term to both sides of $P(k)$ D add the $k$ th term to both sides of $P(k)$

Answer and solution

Answer: (C) add the $(k+1)$ th term to both sides of $P(k)$

Assume $P(k)$ , add the next term, the $(k+1)$ th, to both sides and simplify to get $P(k+1)$ .
To prove $\displaystyle 2 \cdot 3 + 3 \cdot 4 + 4 \cdot 5 + \dots + (n + 1)(n + 2) = \frac{n (n^{2} + 6 n + 11)}{3}$ by induction, the term added to both sides of $P(k)$ is
A $(k + 1) (k + 2)$ B $(k + 3) (k + 4)$ C $(k + 1) (k + 2) + 1$ D $(k + 2) (k + 3)$

Answer and solution

Answer: (D) $(k + 2) (k + 3)$

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

It is $(k + 2) (k + 3)$ .
By induction it can be shown that $7 + 17 + 27 + \dots + (10 n - 3) =$
A $(n - 1) (5 n - 3)$ B $(n + 1) (5 n + 7)$ C $n (5 n + 2)$ D $2 n (5 n + 2)$

Answer and solution

Answer: (C) $n (5 n + 2)$

Check with $n = 1$ : the sum is $7$ and the formula gives $7$ .

The formula $n (5 n + 2)$ is the one proved by induction.
$1 + 2 + 3 + \dots + 66 =$
A $2211$ B $2278$ C $4422$ D $2145$

Answer and solution

Answer: (A) $2211$

$\displaystyle \frac{n(n+1)}{2} = \frac{66 \cdot 67}{2} = 2211$
If $1^3 + 2^3 + \dots + n^3 = 14400$ , then $1 + 2 + \dots + n =$
A $135$ B $15$ C $120$ D $7200$

Answer and solution

Answer: (C) $120$

$\displaystyle \sum n^3 = \left(\sum n\right)^2$

$\displaystyle \therefore \sum n = \sqrt{14400} = 120$
The sum of the first $41$ odd natural numbers is
A $1680$ B $1681$ C $1764$ D $1722$

Answer and solution

Answer: (B) $1681$

$1 + 3 + \dots + (2n - 1) = n^2 = 1681$
The sum of the first $27$ even natural numbers is
A $702$ B $1458$ C $729$ D $756$

Answer and solution

Answer: (D) $756$

$2 + 4 + \dots + 2n = n(n+1) = 27 \cdot 28 = 756$
$1^2 - 2^2 + 3^2 - \dots + 21^2 =$
A $231$ B $-231$ C $462$ D $-210$

Answer and solution

Answer: (A) $231$

$n = 21$ is odd.

$\displaystyle \frac{21 \cdot 22}{2} = 231$
$\displaystyle \sum_{k=1}^{n} k (2 k - 3) =$
A $\displaystyle \frac{n (n - 1) (4 n - 11)}{6}$ B $\displaystyle \frac{n (n + 1) (4 n - 7)}{6}$ C $\displaystyle \frac{n (n + 1) (4 n - 7)}{3}$ D $\displaystyle \frac{(n + 1) (n + 2) (4 n - 3)}{6}$

Answer and solution

Answer: (B) $\displaystyle \frac{n (n + 1) (4 n - 7)}{6}$

Expand $k (2 k - 3) = 2 k^{2} - 3 k$ and use $\displaystyle \sum k^2$ , $\displaystyle \sum k$ .

The total factorises as $\displaystyle \frac{n (n + 1) (4 n - 7)}{6}$ .
$\displaystyle \sum_{k=1}^{10} k^{2} - k + 1 =$
A $910$ B $451$ C $249$ D $340$

Answer and solution

Answer: (D) $340$

Expand $k^{2} - k + 1 = k^{2} - k + 1$ and use the standard sums with $n = 10$ .

The sum is $340$ .
The $15$ th term of the A.P. $1, 11, 21, \dots$ is
A $150$ B $151$ C $131$ D $141$

Answer and solution

Answer: (D) $141$

$t_n = a + (n-1)d = 1 + 14(10) = 141$
The $5$ th term of the G.P. $4, 16, 64, \dots$ is
A $4096$ B $256$ C $1024$ D $65536$

Answer and solution

Answer: (C) $1024$

$t_n = a r^{n-1} = 4 \cdot 4^{4} = 1024$
The sum of the first $6$ terms of $3 + 9 + 27 + \dots$ is
A $3279$ B $1092$ C $2187$ D $2184$

Answer and solution

Answer: (B) $1092$

$\displaystyle S_n = a\frac{r^n - 1}{r - 1} = 3 \cdot \frac{729 - 1}{2} = 1092$
The sum to infinity of $\displaystyle 3 + 7 \cdot \frac{1}{3} + 11 \cdot \frac{1}{9} + 15 \cdot \frac{1}{27} + \dots$ is
A $\displaystyle \frac{15}{2}$ B $\displaystyle \frac{13}{2}$ C $\displaystyle \frac{9}{2}$ D $\displaystyle \frac{27}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{15}{2}$

$a = 3$ , $d = 4$ , $\displaystyle r = \frac{1}{3}$ .

$\displaystyle S_\infty = \frac{a}{1-r} + \frac{dr}{(1-r)^2} = \frac{9}{2} + 3 = \frac{15}{2}$
The $n$ th term of $4, 9, 18, 31, \dots$ is
A $2 n^{2} - n + 3$ B $2 n^{2} - n$ C $2 n^{2} + 3 n + 4$ D $3 n^{2} - 2 n + 2$

Answer and solution

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

Differences: $5, 9, 13$ .

Fitting $n = 1, 2, 3$ gives $T_n = 2 n^{2} - n + 3$ .
The $n$ th term of $10, 46, 190, 766, \dots$ is
A $3 \cdot 4^{n + 1} - 2$ B $3 \cdot 4^{n}$ C $3 \cdot 4^{n} - 2$ D $3 \cdot 4^{n} + 2$

Answer and solution

Answer: (C) $3 \cdot 4^{n} - 2$

Differences: $36, 144, 576$ .

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

Answer and solution

Answer: (D) $(n + 2) (n + 3)$

$T_1 = 12$ , $T_2 = 20$ .

Only $(n + 2) (n + 3)$ gives both.
Testing with $n = 1, 2, 3$ , the sum of $n$ terms of $5 + 12 + 19 + \dots$ is
A $\displaystyle \frac{n (7 n + 3)}{2}$ B $\displaystyle \frac{7 n^{2} - 11 n + 14}{2}$ C $n (7 n - 2)$ D $\displaystyle \frac{(n + 1) (7 n - 2)}{2}$

Answer and solution

Answer: (A) $\displaystyle \frac{n (7 n + 3)}{2}$

$S_1 = 5$ , $S_2 = 17$ , $S_3 = 36$ .

Only $\displaystyle \frac{n (7 n + 3)}{2}$ gives all three.
The largest number that divides $19^{n} - 7^{n}$ for every $n \in \mathbb{N}$ is
A $156$ B $13$ C $12$ D $24$

Answer and solution

Answer: (C) $12$

$n = 1:\ 12,\ n = 2:\ 312$

$\text{G.C.D.}(12, 312) = 12$
The largest number that divides $8^{2 n} - 1$ for every $n \in \mathbb{N}$ is
A $126$ B $63$ C $64$ D $189$

Answer and solution

Answer: (B) $63$

$n = 1:\ 63,\ n = 2:\ 4095$

$\text{G.C.D.}(63, 4095) = 63$
To prove $14^n - 5^n$ is divisible by $9$ , assume $14^k - 5^k = 9\lambda$ . Then $14^{k+1} - 5^{k+1} =$
A $9(\lambda + 5^k)$ B $9(14\lambda + 5^k)$ C $9(5\lambda + 14^k)$ D $9(14\lambda - 5^k)$

Answer and solution

Answer: (B) $9(14\lambda + 5^k)$

$14^{k+1} - 5^{k+1} = 14(9\lambda + 5^k) - 5 \cdot 5^k$

$= 126\lambda + 9 \cdot 5^k = 9(14\lambda + 5^k)$
If $61 \equiv r \pmod{14}$ with $0 \le r < 14$ , then $r =$
A $4$ B $6$ C $9$ D $5$

Answer and solution

Answer: (D) $5$

$61 = 14 \times 4 + 5$

$\therefore 61 - 5 = 56$ is divisible by $14$ , so $r = 5$ .
If $113 \equiv r \pmod{6}$ with $0 \le r < 6$ , then $r =$
A $18$ B $1$ C $0$ D $5$

Answer and solution

Answer: (D) $5$

$113 = 6 \times 18 + 5$

$\therefore 113 - 5 = 108$ is divisible by $6$ , so $r = 5$ .
If $145 \equiv r \pmod{12}$ with $0 \le r < 12$ , then $r =$
A $1$ B $11$ C $12$ D $2$

Answer and solution

Answer: (A) $1$

$145 = 12 \times 12 + 1$

$\therefore 145 - 1 = 144$ is divisible by $12$ , so $r = 1$ .

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MCQ test, Set 9