Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 12

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Question 1
1 / -0

For all $$n\in N, \sum n$$ is

A
$$\displaystyle < \frac{(2n + 1)^2}{8}$$

B
$$\displaystyle > \frac{(2n + 1)^2}{8}$$

C
$$\displaystyle = \frac{(2n + 1)^2}{8}$$

D
none of these

Solution

$$\displaystyle \sum n = \dfrac{n(n+1)}2=\dfrac{n^2+n}2=\dfrac{4n^2+4n}{8}$$

$$\quad = \dfrac{4n^2+4n+1-1}{8}=\dfrac{(2n+1)^2-1}{8}$$

$$\quad =\dfrac{(2n+1)^2}{8}-\dfrac{1}{8}<\dfrac{(2n+1)^2}{8}$$

Hence option 'A' is correct choice.
Question 2
1 / -0

If $$49^{n} + 16 n + \lambda$$ is divisible by $$64$$ for all $$n\in N$$, then the least negative integral value of $$\lambda$$ is

A
$$-12$$

B
$$-1$$

C
$$-3$$

D
$$-4$$

Solution

For $$n = 1$$, we have
$$49^n + 16 n + \lambda = 49 + 16 + \lambda = 65 + \lambda= 64 + (\lambda + 1)$$, which is divisible by 64 if $$\lambda = -1$$.
For $$n = 2$$, we have
$$49^n + 16 n + \lambda = 49^2 + 16 \times 2 + \lambda = 2433 + \lambda= (64 \times 38) + (\lambda + 1)$$, which is divisible by 64 if $$\lambda = -1$$
Hence, $$\lambda = -1$$.
Question 3
1 / -0

Let $$P(m)$$ be the statement $$m^{2}> 100$$, the statement $$P(k + 1)$$ will be true if

A
$$P(1)$$ is true

B
$$P(2)$$ is true

C
$$P(k)$$ is true

D
none of these

Solution

$$P(r)$$ is true
$$\displaystyle \Rightarrow r^{2}> 100$$ $$\displaystyle \Rightarrow r^{2}+2r+1> 100+2r+1$$
$$\displaystyle \Rightarrow \left ( r+1 \right )^{2}> 100$$
$$\displaystyle \Rightarrow P(r+1)$$ is true as
$$\displaystyle r^{2}+\left ( 2r+1 \right )>r^{2}> 100 $$ $$\displaystyle \Rightarrow P\left (k+1 \right )$$ is true (say $$r=k$$)
$$P(k+1)$$ is true when every $$p(k)$$ is
So, In order to prove that $$P(k+1)$$ is true.
It is sufficient to consider $$P(k)$$ is true.
Question 4
1 / -0

Let $$P(n)$$ be a statement such that truth of $$P\left ( n \right )\Rightarrow $$ the truth of $$P\left ( n+1 \right )$$ for all $$n\epsilon N$$, then $$P(n)$$ is true

A
$$\forall n> 1$$

B
$$\forall n$$

C
nothing can be said

D
$$\forall n> k$$ (k is some fixed positive integer)

Solution

Let $$P\left( n \right) $$ be a statement such that truth of $$P\left( n \right) \Rightarrow P\left( n+1 \right) $$ for all $$n\in N$$

Now from the statement of the Principle of Mathematical Induction,

we know that if a statement is assumed true for $$n=k$$ and then if it holds true for $$n=k+1$$ and $$n=1$$, then the statement is true for all $$n$$.

So $$P\left( n \right) \Rightarrow P\left( n+1 \right) $$ is given so for $$P\left( n \right) $$ to be true for all $$n\in N$$, $$P\left( 1 \right) $$ should be true.

But nothing has been stated about $$P\left( 1 \right) $$. so nothing can be said about the truth of $$P\left( n \right) $$.
Question 5
1 / -0

$$\displaystyle x^{3^{n}}+y^{3^{n}}$$ is divisible by $$x+y$$, if

A
$$n$$ is any integer $$\geq0$$

B
$$n$$ is an odd positive integer

C
$$n$$ is an even positive integer

D
$$n$$ is a rational number

Solution
Question 6
1 / -0

For $$n \in N, x^{n + 1} + (x + 1)^{2n - 1}$$ is divisible by

A
$$x$$

B
$$x + 1$$

C
$$x^2 + x + 1$$

D
$$x^2 - x + 1$$

Solution

For $$n = 1$$, we have
$$x^{n + 1} + (x + 1)^{2n - 1}$$
$$= x^2 + (x + 1) = x^2 + x + 1$$, which is divisible by $$x^2 + x + 1.$$
For $$n = 2$$, we have
$$x^{n + 1} + (x + 1)^{2n - 1}$$
$$= x^3 + (x + 1)^3$$
$$= (2x + 1) (x^2 + x + 1)$$, which is divisible by $$x^2 + x + 1.$$
Hence, option (c) is true.
Question 7
1 / -0

Statement 1 : For each natural number $$n, (n + 1)^7 - n^7 - 1$$ is divisible by 7.
Statement 2 : For each natural $$n$$, $$n^7 - n$$ is divisible by 7.

A
A) Statement - 1 is True, Statement - 2 is True; Statement -2 is a correct explanation for Statement - 1

B
B) Statement - 1 is True, Statement - 2 is True; Statement -2 is NOT a correct explanation for Statement - 1

C
C) Statement - 1 is True, Statement-2 is False

D
D) Statement - 1 is False, Statement-2 is True

Solution

It can be checked by using the Principle of mathematical induction that statement - 2 is true for all $$n \in N.$$

So, statement - 2 is correct.

Now, $$(n + 1)^7 - n^7 - 1 = \left \{ (n + 1)^7 - (n + 1) \right \} - \left \{ n^7 - n \right \}$$

Here both the terms are divisible by $$7.$$

Therefore, $$(n + 1)^7 - n^7 - 1$$ is also divisible by $$7$$.

So, statement - 1 is correct and statement-2 is a correct explanation for statement-1.
Question 8
1 / -0

Let $$P\left ( n \right )=n\left ( n+1 \right )$$ is an even number, then which of the following satisfy $$P(n)$$

A
$$P(3)$$

B
$$P(100)$$

C
$$P(50)$$

D
All of these

Solution

Given, $$P(n)=n(n+1)$$

$$\displaystyle \therefore P(3)=3.4=12(even) $$

$$\displaystyle P(100)=100\times 101=10100(even)$$

$$\displaystyle P(50)=50\times 51=2520(even)$$

As $$P(3), P(100),P(50)$$ are even numbers.

Short cut Method: $$P(n)=n(n+1)$$

Product of two consecutive integer (natural numbers) is always even.
Question 9
1 / -0

For each natural number, the statement $$P\left ( n \right )=2^{3n}-1$$ is divisible by

A
$$10$$

B
$$6$$

C
$$7$$

D
None of these.

Solution

$$P\left( n \right) =2^{ 3n }-1=8^n-1$$

$$P(n)={ \left( 1+7 \right) }^{ n }-1$$

$$\Rightarrow P\left( n \right) =1+^n{ { C }_{ 1 } } \times 7+^{ n }{ { C }_{ 2 } }\times 7^{ 2 }+...+^{ n }{ { C }_{ n } }\times 7^{ n }-1$$

$$\Rightarrow P\left( n \right) =7\left( ^{ n }{ { C }_{ 1 } }+^{ n }{ { C }_{ 2 } }7+...+^{ n }{ { C }_{ n } }7^{ n-1 } \right) $$

Therefore, $$P(n)$$ is divisible by $$7$$.
Question 10
1 / -0

Let $$P\left ( n \right ):n^{2}+n$$ is an odd integer
$$P\left ( k \right )\Rightarrow P\left ( k+1 \right )$$ is true
Then $$P\left ( n \right )$$ is true for all

A
$$n> 2$$

B
$$n> 1$$

C
$$n$$

D
none of these

Solution

Given that $$P\left( n \right) :{ n }^{ 2 }+n$$ is an odd integer.

$$P\left( k \right) \Rightarrow P\left( k+1 \right) $$ is true.

For $$P\left( n \right) $$ to be true for all $$n$$, it should be true for $$n=1$$

$$P\left( 1 \right) ={ 1 }^{ 2 }+1=2$$ This is not an odd integer, So not true for $$n=1$$

Check for $$n=2$$, $$P\left( 2 \right) ={ 2 }^{ 2 }+2=6$$ This is again not odd,

so not true for $$n=2$$

For $$n=3$$, $$P\left( 3 \right) ={ 3 }^{ 2 }+3=9$$, odd so true

For $$n=4$$, $$P\left( 4 \right) ={ 4 }^{ 2 }+4=20$$ not odd, so not true.

So we see from options, it not true for any given option.

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Principle of Mathematical Induction Test - 12

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Principle of Mathematical Induction Test - 12

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Question 1

$$\displaystyle < \frac{(2n + 1)^2}{8}$$

$$\displaystyle > \frac{(2n + 1)^2}{8}$$

$$\displaystyle = \frac{(2n + 1)^2}{8}$$

none of these

Solution

Question 2

$$-12$$

$$-1$$

$$-3$$

$$-4$$

Solution

Question 3

$$P(1)$$ is true

$$P(2)$$ is true

$$P(k)$$ is true

none of these

Solution

Question 4

$$\forall n> 1$$

$$\forall n$$

nothing can be said

$$\forall n> k$$ (k is some fixed positive integer)

Solution

Question 5

$$n$$ is any integer $$\geq0$$

$$n$$ is an odd positive integer

$$n$$ is an even positive integer

$$n$$ is a rational number

Solution

Question 6

$$x$$

$$x + 1$$

$$x^2 + x + 1$$

$$x^2 - x + 1$$

Solution

Question 7

A) Statement - 1 is True, Statement - 2 is True; Statement -2 is a correct explanation for Statement - 1

B) Statement - 1 is True, Statement - 2 is True; Statement -2 is NOT a correct explanation for Statement - 1

C) Statement - 1 is True, Statement-2 is False

D) Statement - 1 is False, Statement-2 is True

Solution

Question 8

$$P(3)$$

$$P(100)$$

$$P(50)$$

All of these

Solution

Question 9

$$10$$

$$6$$

$$7$$

None of these.

Solution

Question 10

$$n> 2$$

$$n> 1$$

$$n$$

none of these

Solution

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