Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

If $$P(n)$$ be the statement $$n(n+1)+1$$ is odd, then which of the following is false?

A
$$P(2)$$

B
$$P(3)$$

C
$$P(4)$$

D
none of these

Solution

$$\displaystyle P\left ( n \right )= n( n+1 )+1$$

$$\displaystyle P\left ( 2 \right )=6+1=7$$

$$\displaystyle P\left ( 3 \right )=3\times 4+1=13$$

$$\displaystyle P\left ( 4 \right )=4\times 5+1=21$$

$$\displaystyle \therefore $$ None of the above is even.

Ans: D
Question 2
1 / -0

Let $$P\left ( n \right )=2^{3n}-7n-1$$ then $$P(n)$$ is divisible by

A
$$63$$

B
$$36$$

C
$$49$$

D
$$25$$

Solution

$$P\left( n \right) =2^{ 3n }-7n-1=-1-7n+{ \left( 1+7 \right) }^{ n }$$

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

$$\Rightarrow P\left( n \right) ={ 7 }^{ 2 }\left( { n }_{ { C }_{ 2 } }+{ n }_{ { C }_{ 3 } }7+...+{ n }_{ { C }_{ n } }7^{ n-2 } \right) $$

Therefore, $$P(n)$$ is divisible by $$49$$.

Ans: 49
Question 3
1 / -0

Let $$P(n)$$ be the statement representing the sum of next three successive natural numbers of $$n$$, $$\forall n\in N$$, then the smallest value of $$n$$ to which $$P(n)$$ is divisible by $$9$$ is

A
$$1$$

B
$$3!$$

C
$$3$$

D
$$9!$$

Solution

As $$\displaystyle P\left ( n \right )=\left ( n+1 \right )+\left ( n+2 \right )+\left ( n+3 \right )$$

$$\displaystyle P \left ( n \right )=3n+3+3$$

$$\displaystyle P \left ( n \right )=3\left ( n+2 \right )$$

$$\displaystyle \therefore P \left ( 1 \right )=3\left ( 3 \right ) = 9$$

Which is divisible by $$9 \displaystyle $$

$$ \therefore $$ least value of $$n$$ is $$1.$$
Question 4
1 / -0

Let $$P\left ( k \right ):2+4+6+...+2k=k\left ( k+1 \right )+2$$, then the statement $$P(m+1)$$ will be true if

A
$$P(1)$$ is true

B
$$P(2)$$ is true

C
$$P(m)$$ is true

D
none of these

Solution

Given $$P\left ( k \right ):2+4+6+...+2k=k\left ( k+1 \right )+2$$

We have $$P(m+1)$$ is true.
$$\Rightarrow 2+4+6+.....+2m+2(m+1)=(m+1)(m+2)+2$$
$$\Rightarrow 2+4+6+.....+2m=(m+1)(m+2)+2-2(m+1)$$
$$\Rightarrow 2+4+6+.....+2m=(m+1)m+2$$
Hence, $$P(m)$$ is true.

Also, $$P(1)$$ is not true as
$$LHS=2$$
$$RHS=2+2=4$$

Also, $$P(2)$$ is not true as
$$LHS=2+4=6$$
$$RHS=2(3)+2=8$$
Question 5
1 / -0

Let $$\displaystyle P\left ( n \right )=1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^{2}}< 2-\frac{1}{n}$$ is true for

A
$$\forall n$$

B
for $$n=1$$

C
$$n=2$$

D
none of these

Solution

For $$n=1$$
$$L.H.S=P(1)=1$$
$$R.H,S=2-1=1.$$
$$ L.H.S=R.H.S.$$
$$\therefore 1$$ can't be $$<1.$$
$$\Rightarrow P\left( 1 \right) $$is not true.
Again for $$n=2$$
$$ \displaystyle L.H.S:P\left( 2 \right) =1+\frac { 1 }{ 4 } =\frac { 5 }{ 2 } .$$
$$ \displaystyle R.H.S: P\left( 2 \right) =2-\frac { 1 }{ 2 } =\frac { 3 }{ 2 } =\frac { 6 }{ 4 } .$$
$$L.H.S\quad P\left( 2 \right) <R.H.S\quad P\left( 2 \right) $$
$$\therefore P\left( 2 \right) $$ is true.
Question 6
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$$P\left ( n \right ):2^{n+2}< 3^{n}$$, is true for

A
$$n\in N$$

B
$$n>3, n\in N$$

C
$$n>2, \forall n\in N$$

D
none of these.

Solution

$$\displaystyle P\left ( n \right ):2^{n+2}< 3^n$$
Let $$\displaystyle n=1 \,\,\,\,P\left ( 1 \right ):2^{3}< 3^{1}$$ i.e. $$\displaystyle P\left ( 1 \right )=8< 3$$ false
Let $$\displaystyle n=2\,\,\,\, P\left ( 2 \right ):2^{4}< 3^{2}$$ i.e. $$\displaystyle P\left ( 2 \right )=16< 9$$ false
Let $$\displaystyle n=3\,\,\,\, P\left ( 3 \right ):2^{5}< 3^{3}$$ or $$\displaystyle P\left ( 3 \right )=32< 27,$$ false
Let $$n=4$$
$$\displaystyle \therefore\,\,\,\, P\left ( 4 \right ):2^{6}< 3^{4}$$ or $$\displaystyle P(4)=64< 81$$ which is true.
$$\displaystyle P\left ( n \right ):2^{n+2}< 3^{n}$$ is true for $$\displaystyle \forall n> 3,n\epsilon N$$
Question 7
1 / -0

Let $$P(n)$$ be the statement that $$n^{2}-n+41$$ is prime, then which of the following is not true ?

A
$$P(2)$$

B
$$P(3)$$

C
$$P(41)$$

D
none of these

Solution

given that $$\displaystyle P\left ( n \right )=n^{2}-n+41$$

$$\displaystyle P\left ( 2 \right )=2^{2}-2+41=43$$ (prime true)

$$\displaystyle P\left ( 3 \right )=3^{2}-3+41=47$$ (prime true)

$$\displaystyle P\left ( 41\right )=41^{2}-41+41=\left ( 41 \right )^{2}$$

$$\displaystyle \therefore P\left ( 41 \right )$$ is not true

Ans: C
Question 8
1 / -0

Let $$P\left ( n \right ):2^{n}> n, \forall n\in N$$ and $$2^{k}> k, \forall n=k$$, then which of the following is true $$\forall k\geq 2$$?

A
$$2^{k}> 5k> 1$$

B
$$2^{k+1}> 2k> k+1$$

C
$$2^{k}> 2\left ( k+1 \right )> k$$

D
None of these.

Solution

$$\quad{ 2 }^{ k }>k$$, $$\forall n=k$$
So, $${ 2 }^{ k+1 }>k+1$$ ...(i)

Also, $$k+k>k+1$$, $$\forall k\ge 2$$
$$\Rightarrow 2k>k+1$$ ...(ii)

And $${ 2 }^{ k+1 }>{ 2 }^{ k }>2k+1$$ ...(iii)
From (i), (ii) and (iii), we get option $$B$$.
Question 9
1 / -0

The inequality, $$P(n)=n!> 2^{n}$$ is true for

A
$$n\geq 4$$

B
$$n> 1$$

C
$$n> 2$$

D
$$\forall n\in N$$

Solution

$$\displaystyle P\left ( n \right )= n!> 2^{n}$$

$$\displaystyle \therefore P\left ( 1 \right )= 1!> 2^{1} $$ (false)

$$\displaystyle P\left ( 2 \right )= 2!> 2^{2}=4$$ (false)

$$\displaystyle P\left ( 3 \right )= 3!> 2^{3}$$ or $$\displaystyle 6> 8$$ (false)

$$\displaystyle P\left ( 4 \right )=4!> 2^{4}$$ or $$\displaystyle 24> 16 $$ (true)

Hence, inequality holds if $$\displaystyle n\geq 4.$$

Ans: $$n \ge 4$$
Question 10
1 / -0

Let $$P\left ( n \right ):a^{n}+b^{n}$$ such that $$a, b$$ are even, then $$p(n)$$ will be divisible by $$a + b$$ if

A
$$n> 1$$

B
$$n$$ is odd

C
$$n$$ is even

D
none of these

Solution

$$\displaystyle P\left ( n \right )=a^{n}+b^{n}\,\,\,\,\forall n \epsilon N.$$

$$n=1$$

$$\displaystyle \therefore p\left ( 1 \right )= a+b $$ which is divisible by $$a+b.$$

$$n=2$$

$$\displaystyle \therefore P\left ( 2 \right )=a^{2}+b^{2} $$ not divisible by $$a+b$$

$$n=3$$

$$\displaystyle \therefore P\left ( 3 \right )=a^{3}+b^{3}=\left ( a+b \right )\left ( a^{2}-ab+b^{2} \right ) $$ which is divisible by $$a+b.$$

with the help of induction, we conclude that p(n) will be divisible by $$a+b$$ if $$n$$ is odd.

Short Cut Method: $$\displaystyle P\left ( n \right )=a^{n}+b^{n}$$and $$a, b \epsilon 2M $$ (even number) Fact: the sum of two odd powers whose bases are even will be always divisible by the sum of their bases

Therefore $$P(n)$$ will be divisible by $$a+b$$ For all $$\displaystyle n \epsilon 2k+1 $$ such that $$k \displaystyle \epsilon N.$$

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Re-Attempt Weekly Quiz Competition

Principle of Mathematical Induction Test - 13

Result

Principle of Mathematical Induction Test - 13

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SHARING IS CARING

Question 1

$$P(2)$$

$$P(3)$$

$$P(4)$$

none of these

Solution

Question 2

$$63$$

$$36$$

$$49$$

$$25$$

Solution

Question 3

$$1$$

$$3!$$

$$3$$

$$9!$$

Solution

Question 4

$$P(1)$$ is true

$$P(2)$$ is true

$$P(m)$$ is true

none of these

Solution

Question 5

$$\forall n$$

for $$n=1$$

$$n=2$$

none of these

Solution

Question 6

$$n\in N$$

$$n>3, n\in N$$

$$n>2, \forall n\in N$$

none of these.

Solution

Question 7

$$P(2)$$

$$P(3)$$

$$P(41)$$

none of these

Solution

Question 8

$$2^{k}> 5k> 1$$

$$2^{k+1}> 2k> k+1$$

$$2^{k}> 2\left ( k+1 \right )> k$$

None of these.

Solution

Question 9

$$n\geq 4$$

$$n> 1$$

$$n> 2$$

$$\forall n\in N$$

Solution

Question 10

$$n> 1$$

$$n$$ is odd

$$n$$ is even

none of these

Solution

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