Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 17

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Question 1
1 / -0

If x > 1, then the statement $$P(n) : (1 + x)^n > 1 + nx$$ is true for

A
all $$n\epsilon N$$

B
all n > 1

C
all n > 1 and $$x\neq 0$$

D
None of these

Solution

If $$x > 1$$ then the statement

$$P(n) : (1 + x)^n > 1 + nx$$

$$P(1) : (1 + x)^1 > 1 + x$$, which is false

$$P(2) : (1 + x)^2 > 1 + 2x$$, which is true if $$x\neq 0$$

Let $$P(k) : (1 + x)^k > 1 + kx$$ is true

$$\Rightarrow (1 + x) (1 + x)^k > (1 + x) (1 + kx)$$

$$\Rightarrow (1 + x)^{k+1} > (1 + x) (1 + kx) = 1 + (k + 1)x + kx^2$$

$$\Rightarrow (1 + x)^{k+1} > 1 + (k + 1)x \because kx^2 > 0$$

Hence $$P(n)$$ is true

so $$P(n)$$ is true for all $$n > 1$$ and $$x\neq 0$$
Question 2
1 / -0

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

A
no $$n\epsilon N$$

B
all $$n\epsilon N$$

C
all $$n\geq 5$$

D
none of these

Solution
Question 3
1 / -0

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

A
n > 1

B
$$n\geq 1$$

C
n > 2

D
never

Solution

Given equation,
$$\left(\dfrac{n+1}{2}\right)^n\ge{n}!$$
For $$n=1$$,
LHS=$$\left(\dfrac{1+1}{2}\right)^1=1$$
RHS=$$1!=1$$
For $$n=2$$,
LHS=$$\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}=2.25$$
RHS=$$2!=2$$
Given condition is true for $$n\ge1$$. Hence, it is the answer.
Question 4
1 / -0

If $$n \in N$$, then $$x^{2n+1} + y^{2n+1}$$ is divisible by

A
$$x+y$$

B
$$x-y$$

C
$$x^2+y^2$$

D
$$x^2+xy$$

Solution
Question 5
1 / -0

The inequality $$n! > 2^{n-1}$$ is true

A
for all $$n > 1$$

B
for all $$n > 2$$

C
for all $$n \epsilon N$$

D
none of these

Solution

Given inequality,
$$n!>2^{n-1}$$
For $$n=1$$,
$$LHS=n!=1!=1$$
$$RHS=2^{n-1}=2^{1-1}=2^0=1$$
Inequality not satisfied.
For $$n=2$$,
$$LHS=n!=2!=2\times1=2$$
$$RHS=2^{n-1}=2^{2-1}=2^1=2$$
Inequality not satisfied.
For $$n=3$$,
$$LHS=n!=3!=3\times2\times1=6$$
$$RHS=2^{n-1}=2^{3-1}=2^2=4$$
Inequality is satisfied in this case.
Hence, the inequation is true for $$n>2$$.
Question 6
1 / -0

For positive integer n, $$10^{n-2} > 81n$$ when

A
$$n < 5$$

B
$$n > 5$$

C
$$n\geq 5$$

D
$$n > 6$$

Solution

$$10^{n-2}>81n$$
For $$n=5$$,
LHS=$$10^{5-2}=10^3=1000$$
RHS=$$81n=81\times5=405$$
$$\Rightarrow10^{n-2}>81n$$ for all $$n\ge5$$. Hence, the answer.
Question 7
1 / -0

State which of the following statements is true $$ \displaystyle 2^{16}-1 $$ is divisible by

A
11

B
13

C
17

D
19

Solution

2$$\displaystyle ^{16}-1=\left ( 2^{4} \right )^{4}-1=\left ( 16 \right )^{4}-1$$
=(16-1)(16+1)(16$$\displaystyle ^{2}$$+1)
Hence 2$$\displaystyle ^{16}$$-1 is divisible by 17
Question 8
1 / -0

Let $$S(K) =1+ 3 + 5...+ (2K -1) =3 + K^2$$. Then which of the following is true

A
Principle of mathematical induction can be used to prove the formula

B
$$S(K) \Rightarrow S(K + 1)$$

C
$$S(K) \not{\Rightarrow} S(K + 1)$$

D
S(1) is correct

Solution

$$S(k) = 1+3+5+...+(2k-1)=3+k^2$$

$$S(1) = 1= 3+ 1,$$ which is not true

$$\because $$ S(1) is not true.

$$\therefore $$ P.M. 1 cannot be applied

Let $$S(k)$$ is true, i.e.

$$1+3+5.... +(2k-1)=3+k^2$$

$$\Rightarrow 1+3+5....+ (2k-1)+2k +1$$

$$= 3+k^2 +2k+I=3+(k+1)^2$$

$$\therefore S(k) \Rightarrow S(k + 1)$$
Question 9
1 / -0

The statement P(n) $$"1 \times 1! + 2\times 2! + 3\times 3! + ... + n \times n! =(n + 1)! 1"$$ is

A
True for all n > 1

B
Not true for any n

C
True for all $$n\epsilon N$$

D
None of these

Solution

$$P\left( 1 \right) :1\times 1=\left( 1+1 \right) =\left( 1+1 \right) !-1$$ is true
Let $$p(m)$$ be true
$$\Rightarrow 1\times 1!+2\times 2!+3\times 3!+...+m\times m!=\left( m+1 \right) !-1$$
$$\Rightarrow 1\times 1+2\times 2!+3\times 3!+...+m\times m!+\left( m+1 \right) \times \left( m+1 \right) !$$
$$=\left( m+1 \right) !-1\left( m+2 \right) \left( m+1 \right) !=\left( m+1 \right) !\left( m+2 \right) -1$$
$$=\left( m+2 \right) !-1$$
$$\Rightarrow p(m+1)$$ is also true
$$\therefore P(n)$$ is true for each $$n$$
Question 10
1 / -0

If $$a, b$$ and $$n$$ are natural numbers then $$a^{2n-1}+b^{2n-1}$$ is divisible by

A
$$a + b$$

B
$$a - b$$

C
$$a^3+b^3$$

D
$$a^2+b^2$$

Solution

Let $$P(n)=a^{2n-1}+b^{2n-1}$$

$$P(1)=a+b$$
$$P(2)=a^3+b^3=(a+b)(a^2+b^2-ab)$$
Out of given options we can conclude that $$P(n)$$ is divisible only by $$(a+b)$$

It can also be done by Mathematical Induction.

But here it is objective, so verification is good enough.

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Principle of Mathematical Induction Test - 17

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Principle of Mathematical Induction Test - 17

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Question 1

all $$n\epsilon N$$

all n > 1

all n > 1 and $$x\neq 0$$

None of these

Solution

Question 2

no $$n\epsilon N$$

all $$n\epsilon N$$

all $$n\geq 5$$

none of these

Solution

Question 3

n > 1

$$n\geq 1$$

n > 2

never

Solution

Question 4

$$x+y$$

$$x-y$$

$$x^2+y^2$$

$$x^2+xy$$

Solution

Question 5

for all $$n > 1$$

for all $$n > 2$$

for all $$n \epsilon N$$

none of these

Solution

Question 6

$$n < 5$$

$$n > 5$$

$$n\geq 5$$

$$n > 6$$

Solution

Question 7

11

13

17

19

Solution

Question 8

Principle of mathematical induction can be used to prove the formula

$$S(K) \Rightarrow S(K + 1)$$

$$S(K) \not{\Rightarrow} S(K + 1)$$

S(1) is correct

Solution

Question 9

True for all n > 1

Not true for any n

True for all $$n\epsilon N$$

None of these

Solution

Question 10

$$a + b$$

$$a - b$$

$$a^3+b^3$$

$$a^2+b^2$$

Solution

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