Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 16

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Question 1
1 / -0

If $$P$$ is a prime number then $$n^{p} - n$$ is divisible by $$p$$ when $$n$$ is a

A
natural number greater than 1

B
odd number

C
even number

D
None of these

Solution

Let $$P(n) :$$ $$n^{p} - n$$

when $$p = 2$$

$$P(n) =$$ $$n^{2} - n$$

$$P(1) = 0$$ which is divisible all n$$\in$$N

$$P(2) = 2$$ which is divisible by $$2$$

$$P(3) = 6$$ which is divisible by $$2$$

Hence $$P(n)$$ is divisible by $$2$$ when $$n$$ is greater than $$1$$
Question 2
1 / -0

The difference between an $$+$$ve integer and its cube, is divisible by

A
$$4$$

B
$$6$$

C
$$9$$

D
None of these

Solution

Let $$n$$ be the positive integer.

Now its cube is $${ n }^{ 3 }$$

We have to check for divisibility of $${ n }^{ 3 }-n$$

$$\Rightarrow { n }^{ 3 }-n=n\left( n-1 \right) \left( n+1 \right) =\left( n-1 \right) n\left( n+1 \right) $$

This clearly is the product of $$3$$ consecutive numbers.

The product of $$3$$ consecutive numbers is divisible by $$6$$.

As per the divisibility rule for $$6,$$ the number should be divisible both by $$2$$ and $$3.$$

In the product of $$3$$ consecutive numbers, the number will turn out to be
divisible by both $$2$$ and $$3.$$

This can be proved by Mathematical Induction.
Question 3
1 / -0

If n is a natural number then $$\left( \displaystyle \frac{n + 1}{2} \right)^{n}\, \geq n!$$ is true when.

A
n > 1

B
$$n \geq 1$$

C
n > 2

D
Never

Solution

If $$n$$ is a natural number and $$ \left(\dfrac{n+1}{2}\right)^n$$
Verifying for $$n=1$$
$$\left(\dfrac{1+1}{2}\right)^1= \left(\dfrac{2}{2}\right)=1\ge{1}!$$
Therefore, the condition is satisfied.
From option $$\left(\dfrac{n+1}{2}\right)^{n}\ge{n}!$$
For $$n\ge1$$ and also $$n=0$$
Hence, option B is the correct answer.
Question 4
1 / -0

For all n $$\in$$ N, $$n^{4}$$ is less than

A
$$10^{n}$$

B
$$4^{n}$$

C
$$10^{10}$$

D
None of these

Solution

For $$n=1$$,
$$n^4=1$$
Option A : 10
Option B: 4
For $$n=2$$,
$$n^4=16$$
Option A: 100
Option B: 16
Thus, $$n$$ is always less than $$10^n$$.
Question 5
1 / -0

A student was asked to prove a statement by induction. He proved
(i) P(5) is true and
(ii) Trutyh of P(n) $$\Rightarrow$$ truth of p(n + 1), n$$\in$$N
On the basis of this, he could conclude that P(n) is true for

A
no n $$\in$$ N

B
all n $$\in$$ N

C
all n $$\geq$$ 5

D
None of these

Solution

If $$P(a)$$ is true and truth of $$P(n)$$ implies that $$P(n+1)$$ is also true, then, we can conclude that $$P(n)$$ is true for all $$n\geq a$$ where $$n\in N$$.
Here
$$P(5)$$ is true.
Hence we can conclude that $$P(n)$$ is true for all $$n\geq 5$$
Question 6
1 / -0

For every natural number $$n$$, $$n(n + 3)$$ is always :

A
multiple of $$4$$

B
multiple of $$5$$

C
even

D
odd

Solution

$$n(n+3)=n^2+3n+2-2=(n+1)(n+2)-2\equiv\text{even}-\text{even}\equiv \text{even}$$

Since for any natural number $$n$$, $$(n+1)(n+2)$$ will always be an even number.

OR

We know there are two types of natural numbers even and odd

Case 1. If $$n$$ is even then $$n+3$$ will be odd

Thus $$n(n+3) = $$ even $$\times $$ odd $$=$$ even

Case 2. If $$n$$ is odd then $$n+3$$ will be even

Thus $$n(n+3) = $$ odd $$\times $$ even $$=$$ even

Hence $$n(n+3)$$ will always be even number
Question 7
1 / -0

For every positive integral value of n, $$3^n > n^3$$ when

A
n > 2

B
$$n\geq 3$$

C
$$n\geq 4$$

D
n < 4

Solution

Let $$P(n) : 3^n > n^3$$

$$P(1) : 3 > 1$$, which is true

$$P(2) : 3^2 > 2^3$$, which is true

$$P(3) : 3^3 > 3^3$$, which is false

$$P(4) : 3^4 > 4^3$$, which is true

$$P(5) : 3^5 > 5^3$$, which is true etc.

Hence $$P(n)$$ is true when $$n > 4.$$
Question 8
1 / -0

$$P(n) : 3^{2n+2} -8n -9$$ is divisible by 64, is true for

A
all $$n\epsilon N \cup \left \{0\right \}$$

B
$$n\geq 2, n\epsilon N$$

C
$$n\epsilon N, n > 2$$

D
none of these

Solution
Question 9
1 / -0

The smallest positive integer for which the statement $$3^{n+1} < 4^n$$ holds is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Let $$P(n) : 3^{n + 1} < 4^n$$

$$P(1) : 3^2 < 4$$ which is false

$$P(2) : 3^3 < 4^2$$ which is false

$$P(3) : 3^4 < 4^3$$ which is false

$$P(4) : 3^5 < 4^4$$ which is true
Question 10
1 / -0

For every positive integer $$n, \dfrac {n^7}{7}+\dfrac {n^5}{5}+\dfrac {2n^3}{3}-\dfrac {n}{105}$$ is

A
an integer

B
a rational number

C
a negative real number

D
an odd integer

Solution

$$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 } $$
where $$n$$ is a positive integer.
When $$n=1$$, we get,
$$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 } $$
$$=\dfrac { 15{ n }^{ 7 }+21{ n }^{ 5 }+70{ n }^{ 3 }-n }{ 105 } $$
$$=\dfrac { 15+21+70-1 }{ 105 } =\dfrac { 105 }{ 105 } =1$$ which is an integer.
When $$n=2$$,
we get;
$$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 } $$
$$=\dfrac { 15{ n }^{ 7 }+21{ n }^{ 5 }+70{ n }^{ 3 }-n }{ 105 } $$
$$=\dfrac { \left( 15\times 128 \right) +\left( 21\times 32 \right) +\left( 70\times 28 \right) -2 }{ 105 } $$
$$=\dfrac { 1920+672+560-2 }{ 105 } $$
$$=\dfrac { 3105 }{ 105 } =30$$ which is an even positive integer.

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Principle of Mathematical Induction Test - 16

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Principle of Mathematical Induction Test - 16

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Question 1

natural number greater than 1

odd number

even number

None of these

Solution

Question 2

$$4$$

$$6$$

$$9$$

None of these

Solution

Question 3

n > 1

$$n \geq 1$$

n > 2

Never

Solution

Question 4

$$10^{n}$$

$$4^{n}$$

$$10^{10}$$

None of these

Solution

Question 5

no n $$\in$$ N

all n $$\in$$ N

all n $$\geq$$ 5

None of these

Solution

Question 6

multiple of $$4$$

multiple of $$5$$

even

odd

Solution

Question 7

n > 2

$$n\geq 3$$

$$n\geq 4$$

n < 4

Solution

Question 8

all $$n\epsilon N \cup \left \{0\right \}$$

$$n\geq 2, n\epsilon N$$

$$n\epsilon N, n > 2$$

none of these

Solution

Question 9

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 10

an integer

a rational number

a negative real number

an odd integer

Solution

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