Principle Of Mathematical Induction(11th Grade > Mathematics ) Questions and answers for exam Preparation

Let P(n) denote the statement that $n^{2}$ + n is odd. Then,

P(1) is true
P(2) is true
$P (k) \Rightarrow P (k + 1)$
$P(n) is true \forall n ϵ N$

Explanation:-

Answer: Option C. ->

P (k) \Rightarrow P (k + 1)

:
C

$P (1) : 1^{2} + 1 is odd \to n o t t r u e P (2) : 2^{2} + 2 is odd \to n o t t r u e$
Suppose P(k) is true.
$k^{2} + k is odd \Rightarrow k^{2} + k = 2 m + 1 P (k + 1) : (k + 1)^{2} + (k + 1) = (k^{2} + 2 k + 1) + (k + 1) = (k^{2} + k) + (2 k + 2) = 2 m + 1 + 2 k + 2 2 (m + k + 1) + 1 \Rightarrow P(k+1) is true$

Question 2.

If n is a natural number then ${(\frac{n + 1}{2})}^{n}$ ≥ n ! is true

when

n > 1
$n \geq$ 1
n > 2
$n \geq 2$

Explanation:-

Answer: Option B. ->

n \geq

1
:
B

Check through option, the condition

${(\frac{n + 1}{2})}^{n}$ ≥ n ! is true for n ≥ 1.

Question 3.

To find:
$1^{2} + 2^{2} + 3^{2} . . . . . . . . . . . . . . . . . . . + n^{2}$

$= \frac{n^{3}}{2}$
$> \frac{n^{3}}{2}$
$> \frac{n^{3}}{3}$
$= \frac{n^{3}}{3}$

Explanation:-

Answer: Option C. ->

> \frac{n^{3}}{3}

:
C
Let

n = 1

\frac{n^{3}}{2} = \frac{1}{2}; \frac{n^{3}}{3} = \frac{1}{3}

Let

n = 2, \Rightarrow 1^{2} + 2^{2} = 5

\frac{n^{3}}{2} = 4; \frac{n^{3}}{3} = \frac{8}{3}

Let

n = 3, \Rightarrow 1^{2} + 2^{2} + 3^{2} = 14

\frac{n^{3}}{2} = \frac{27}{2}; \frac{n^{3}}{3} = 9

Let

n = 4, \Rightarrow 1^{2} + 2^{2} + 3^{2} + 4^{2} = 30

\frac{n^{3}}{2} = 32; \frac{n^{3}}{3} = \frac{64}{3}

1^{2} + 2^{2} + 3^{2} + \dots + n^{2} > \frac{n^{3}}{3}

P (n) : 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} > \frac{n^{3}}{3}

P (1)

is true
Let

P (k)

be true.

\Rightarrow 1^{2} + 2^{2} + 3^{2} + \dots + k^{2} > \frac{k^{3}}{3}

1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + (k + 1)^{2} > \frac{k^{3}}{3} + k^{2} + 2 k + 1

= \frac{k^{3} + 3 k^{2} + 6 k + 3}{3}

= \frac{k^{3} + 3 k^{2} + 3 k + 1 + 3 k + 2}{3}

= \frac{(k + 1)^{3}}{3} + k + \frac{2}{3}

1^{2} + 2^{2} + 3^{2} + \dots + k^{2} + (k + 1)^{2} > \frac{(k + 1)^{3}}{3}

P (k + 1)

is true.

P (n)

is true

\forall n \in N

Question 4.

$P (n) : 5^{2 n + 1} + 3^{n + 2} {.2}^{n - 1}$ is divisible by

19
18
17
14

Explanation:-

Answer: Option A. -> 19
:
A

$P (n) : 5^{2 n + 1} + 3^{n + 2} {.2}^{n - 1}$
$P (1) : 5^{3} + 3^{3} .1 = 152 \to divisible by 19$
Assume P(k) is true.
$5^{2 k + 1} + 3^{k + 2} {.2}^{k - 1} is divisible by 19 5^{2 k + 1} + 3^{k + 2} {.2}^{k - 1} = 19 m P (k + 1) : 5^{2 k + 3} + 3^{k + 3} {.2}^{k} = {25.5}^{2 k + 1} + {6.3}^{k + 2} {.2}^{k - 1} = 6. (5^{2 k + 1} + 3^{k + 2} {.2}^{k - 1}) + {19.5}^{2 k + 1} = 19 (6 m + 5^{2 k + 1})$

Question 5.

$P (n) : 1 + 3 + 3^{2} + . . . + 3^{n - 1} = \frac{3^{n} - 1}{2}$
The statement P(n)

is true for all natural numbers
is not true for all natural numbers
is true only for a finite number of natural numbers
is true only for all natural numbers greater than 2

Explanation:-

Answer: Option A. -> is true for all natural numbers
:
A

$P (n) : 1 + 3 + 3^{2} + . . . + 3^{n - 1} = \frac{3^{n} - 1}{2} P (1) : 1 = \frac{3 - 1}{2} \Rightarrow 1 = 1 - t r u e$
Assume P(k) is true
$\Rightarrow 1 + 3 + 3^{2} + . . . + 3^{k - 1} = \frac{3^{k} - 1}{2} N o w, P (k + 1) : 1 + 3 + 3^{2} + . . . + 3^{k - 1} + 3^{k} = \frac{3^{k} - 1}{2} + 3^{k} = 3^{k} (1 + \frac{1}{2}) - \frac{1}{2} = \frac{3^{k + 1} - 1}{2}$
P(k+1) is also true.
Hence, P(n) is true for all natural numbers

Question 6.

'For all natural numbers N, if P(n) is a statement about n and P(k+1) is true if P(k) is true for an arbitrary natural number k, then P(n) is always true.' State true or false.

True
False
is true only for a finite number of natural numbers
is true only for all natural numbers greater than 2

Explanation:-

Answer: Option B. -> False
:
B

For the proof by mathematical induction to work, the statement P(n) must be true for a specific instance of a natural number.
Hence, if P(m) is true, where m is a specific natural number and P(k+1) is true if P(k) is true for an arbitrary natural number k, then, $P (n) i s t r u e \forall n \geq m$
Without the base case P(m), we cannot say that P(n) is true. Hence, the statement is false.

Question 7.

'If P(n) is a statement about natural numbers, mathematical induction may be used to prove P(n) only if P(1) is true.' State true or false.

True
False
is true only for a finite number of natural numbers
is true only for all natural numbers greater than 2

Explanation:-

Answer: Option B. -> False
:
B

The principle of mathematical induction can be used to prove true statements P(n) about natural numbers if it is true for a specific natural number, 'm.' In this case, if proved using induction, P(n) will be true for all $n \geq m .$ 'm' need not be equal to 1.

Question 8.

$P (n) : 1 + 3 + 5 + . . . + 2 n - 1 = n^{2}$
The statement P(n) is

is true for all natural numbers
is not true for n>1
is true for n>2
none of these

Explanation:-

Answer: Option A. -> is true for all natural numbers
:
A

$P (n) : 1 + 3 + 5 + . . . + 2 n - 1 = n^{2} P (1) : 1 = 1^{2} \Rightarrow 1 = 1 - t r u e$
Assume P(k) is true
$1 + 3 + 5 + . . . + 2 k - 1 = k^{2} N o w, P (k + 1) : 1 + 3 + 5 + . . . + 2 n - 1 + 2 k + 1 = k^{2} + 2 k + 1 = (k + 1)^{2}$
P(k+1) is also true.
Hence, P(n) is true for all natural numbers

Question 9.

$P (n) : 1.2 + 2.3 + 3.4 + . . . + n (n + 1) = \frac{(n + 1) (n + 2)}{3}$
The statement P(n) is

is true for all natural numbers
is not true for all natural numbers
is true for n=1
is true for n=2

Explanation:-

Answer: Option B. -> is not true for all natural numbers
:
B

$P (n) : 1.2 + 2.3 + 3.4 + . . . + n (n + 1) = \frac{(n + 1) (n + 2)}{3} P (1) : 1.2 = \frac{2.3}{3} \Rightarrow 2 = 2 - t r u e$
Assume P(k) is true
$\Rightarrow 1.2 + 2.3 + 3.4 + . . . + k (k + 1) = \frac{(k + 1) (k + 2)}{3} N o w, P (k + 1) : 1.2 + 2.3 + 3.4 + . . . + k (k + 1) + (k + 1) (k + 2) = \frac{(k + 1) (k + 2)}{3} + (k + 1) (k + 2) = \frac{4 (k + 1) (k + 2)}{3} \neq \frac{(k + 2) (k + 3)}{3}$
P(k+1) is not true.
Hence, P(n) is not true for all natural numbers

Question 10.

$P (n) : a^{2 n} - b^{2 n} is divisible by a + b, \forall n ϵ N$

To prove P(n) using mathematical induction, the base case is

$a^{2} - b^{2} = (a + b) (a - b) is divisible by a + b$
$a^{2 k} - b^{2 k} is divisible by a + b$
$a^{k} - b^{k} is divisible by a + b$
$a^{1} - b^{1} is divisible by a + b$

Explanation:-

Answer: Option A. ->

a^{2} - b^{2} = (a + b) (a - b) is divisible by a + b

:
A

$P (n) : a^{2 n} - b^{2 n} is divisible by a + b, \forall n ϵ N$

Since we need to prove P(n) for all natural numbers, the base case will be n=1.
Substituting n=1, we get
$a^{2} - b^{2} = (a + b) (a - b) is divisible by a + b$