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

444

666

999

None of these

Solution

Question 3

n > 1

n≥1n \geq 1n≥1

n > 2

Never

Solution

Question 4

10n10^{n}10n

4n4^{n}4n

101010^{10}1010

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 444

multiple of 555

even

odd

Solution

Question 7

n > 2

n≥3n\geq 3n≥3

n≥4n\geq 4n≥4

n < 4

Solution

Question 8

all nϵN∪{0}n\epsilon N \cup \left \{0\right \}nϵN∪{0}

n≥2,nϵNn\geq 2, n\epsilon Nn≥2,nϵN

nϵN,n>2n\epsilon N, n > 2nϵN,n>2

none of these

Solution

Question 9

111

222

333

444

Solution

Question 10

an integer

a rational number

a negative real number

an odd integer

Solution

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$4$

$6$

$9$

$n \geq 1$

$10^{n}$

$4^{n}$

$10^{10}$

no n $\in$ N

all n $\in$ N

all n $\geq$ 5

multiple of $4$

multiple of $5$

$n\geq 3$

$n\geq 4$

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

$n\geq 2, n\epsilon N$

$n\epsilon N, n > 2$

$1$

$2$

$3$

$4$