Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 14

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Question 1
1 / -0

Let $$f\left ( n \right ) = 8^{n}-3^{n}$$, if $$n$$ is odd natural number then $$f\left ( n \right )$$ is divisible by

A
$$2$$

B
$$3$$

C
$$5$$

D
none of these

Solution

$$f\left ( n \right ) = 8^{n}-3^{n}$$
Since, $$n$$ is odd
for $$n=1$$, we get $$f\left ( 1 \right ) = 8^{1}-3^{1}=5$$
for $$n=3$$, we get $$f\left ( 3 \right ) = 8^{3}-3^{3}=5(97)$$
for $$n=5$$, we get $$f\left ( 3 \right ) = 8^{5}-3^{5}=5(6505)$$
Therefore, by induction we can say that $$f(n)$$ is divisible by $$5$$ for odd $$n$$.

Ans: C
Question 2
1 / -0

If $$\displaystyle P\left ( n \right )= 1^{2}+3^{2}+5^{2}+...+\left ( 2n-1 \right )^{2}= \frac{n\left ( 4n^{2} -1\right )}{3}$$, then which of the following does NOT hold good?

A
$$P\left ( 1 \right )$$

B
$$P\left ( 2 \right )$$

C
$$P\left ( 3 \right )$$

D
None of these.

Solution

We are given $$P\left( n \right) ={ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }+......{ \left( 2n-1 \right) }^{ 2 }=\displaystyle\frac { n\left( 4{ n }^{ 2 }-1 \right) }{ 3 } $$
We check the validity of this for the given options.
A. $$P\left( 1 \right) $$
LHS$$={ 1 }^{ 2 }=1$$, RHS$$=\displaystyle\frac { 1\left( 4{ \left( 1 \right) }^{ 2 }-1 \right) }{ 3 } =1$$
So True for $$n=1$$.
B. $$P\left( 2 \right) $$
LHS$$=P\left( 2 \right) ={ 1 }^{ 2 }+{ 3 }^{ 2 }=10$$, RHS$$=\displaystyle\frac { 2\left( 4{ \left( 2 \right) }^{ 2 }-1 \right) }{ 3 } =10$$
True for $$n=2$$
C. $$P\left( 3 \right) $$
LHS$$=P\left( 3 \right) ={ 1 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 }=35$$, RHS$$=\displaystyle\frac { 3\left( 4{ \left( 3 \right) }^{ 2 }-1 \right) }{ 3 } =35$$
True for $$n=3$$.
So ,it is true for $$n=1,2,3$$ so NOT true is option D
Question 3
1 / -0

If $$P\left ( n \right )= 1+2+3+\dots+n$$ is a perfect square, $$N$$ is less than $$100$$, then possible values of $$n$$ is/are

A
only $$1$$

B
$$1$$ and $$8$$

C
only $$8$$

D
$$1, 8, 49$$

Solution

$$P(n)\quad =\sum _{ 1 }^{ n }{ { r } =\dfrac {n\times (n+1)}{2}\quad } $$

$$P(1) = 1$$

$$P(8) = 36 =$$ square of $$6$$

$$P(49) = 1225 =$$ square of $$35$$
Question 4
1 / -0

Let the statement $$P\left ( n \right ):2^{n}\geq 3n$$, the truth of $$P\left( k \right), \forall k\in N \Rightarrow$$ truth of,

A
$$P\left ( 1 \right )$$

B
$$P\left ( 2 \right )$$

C
$$P\left ( 3 \right )$$

D
$$P\left ( k+1 \right )$$

Solution

Given $$P\left ( n \right ):2^{n}\geq 3n$$ .....(1)

Consider $$P\left ( k+1 \right )$$

$$\Rightarrow 2^{k+1}\ge 3k+3$$

$$\Rightarrow 2^k . 2 \ge 3k+3$$

$$\Rightarrow 2^k \ge \dfrac{3k+3}{2} \ge 3k$$

Hence, $$2^{k} \ge 3k$$

$$\text{Truth of}\,\, P(k+1) \Rightarrow \text{Truth of P(k)}$$

Also, we will check for $$P(1),P(2),P(3)$$

$$P(1): 2 \ngeq 3$$

So, $$P(1)$$ is not true

$$P(2): 4 \ngeq 6$$

So,$$P(2)$$ is not true

$$P(3): 8 \ngeq 9$$

So,$$P(3)$$ is not true
Question 5
1 / -0

Let $$f\left ( n \right )$$ equals to the sum of the cubes of three consecutive natural numbers.
$$f\left ( n \right )$$ leaves the remainder zero when divided by

A
$$11$$

B
$$9$$

C
$$99$$

D
none of these

Solution

Given that $$f\left( n \right) ={ \left( n-1 \right) }^{ 3 }+{ n }^{ 3 }+{ \left( n+1 \right) }^{ 3 }=3n^{ 3 }+6n$$
Put $$n=1$$, to obtain $$f\left( 1 \right) =3.1^{ 3 }+6.1=9$$
Therefore, $$f\left( 1 \right)$$ is divisible by $$9$$
Assume that for $$n=k$$, $$f\left( k \right)=3k^{ 3 }+6k$$ is divisible by $$9$$
Now, $$f\left( k+1 \right) =3\left( k+1 \right) ^{ 3 }+6\left( k+1 \right) =3{ k }^{ 3 }+6k+9\left( { k }^{ 2 }+k+1 \right) =f\left( k \right) +9\left( { k }^{ 2 }+k+1 \right) $$
Since, $$f\left( k \right) $$ is divisible by $$9$$
Therefore, $$f\left( k+1 \right) $$ is divisible by $$9$$
And from the principle of mathematical induction $$f\left( n \right) $$ is divisible by $$9$$ for all $$n\in N$$
Hence, option B is correct.
Question 6
1 / -0

The number of values of $$n$$, for which $$\displaystyle p(n)=1!\:+2!\:+3!\:+4!\:+\dots+\:n!$$ is the square of a natural number, is equal to

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

For $$n = 4$$, $$P(n) = 1+2+6+24 = 33$$.
For $$n >4$$, $$n!$$ will always have $$0$$ in the units place.
$$3 $$ will be unit place digit of p(n) hence can not be sqaure of any natural no.
So the $$n=1,3$$ are the answers.
Question 7
1 / -0

If $$n\in N$$, then $$n^{3}+2n$$ is divisible by

A
$$2$$

B
$$3$$

C
$$5$$

D
$$6$$

Solution

$$f\left( n \right) =n^{ 3 }+2n$$
put $$n=1$$, to obtain $$f\left( 1 \right) =1^{ 3 }+2.1=3$$
Therefore, $$f\left( 1 \right)$$ is divisible by $$3$$
Assume that for $$n=k$$, $$f\left( k \right)=k^{ 3 }+2k$$ is divisible by $$3$$
Now, $$f\left( k+1 \right) =\left( k+1 \right) ^{ 3 }+2\left( k+1 \right) ={ k }^{ 3 }+2k+3\left( { k }^{ 2 }+k+1 \right) =f\left( k \right) +3\left( { k }^{ 2 }+k+1 \right) $$
Since, $$f\left( k \right) $$ is divisible by $$3$$
Therefore, $$f\left( k+1 \right) $$ is divisible by $$3$$
and from the principle of mathematical induction $$f\left( n \right) $$ is divisible by $$3$$ for all $$n\in N$$

Ans: B
Question 8
1 / -0

If $$\displaystyle P\left ( n \right )=\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$$, then statement "$$P(n)$$ is a natural number" is true,

A
only for $$n> 1$$.

B
only for $$n$$ is an odd positive integer.

C
only for $$n$$ is an even positive integer.

D
$$n\in N$$

Solution

Given $$\displaystyle P\left ( n \right )=\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$$

for $$n=1$$ $$\displaystyle P\left ( 1 \right )=\frac{1}{5}+\frac{1}{3}+\frac{7}{15}=\frac{15}{15} = 1$$ (true)

for $$n=2$$ $$\displaystyle P\left (2 \right )=\frac{32}{5}+\frac{8}{3}+\frac{14}{15}=\frac{150}{15} =10$$ (true )

$$\displaystyle \therefore P\left (n \right )$$ is true for $$n=2$$

Similarly by induction we say that $$P(n)$$ is a natural number $$\displaystyle \forall n \displaystyle \epsilon N$$

Ans: D
Question 9
1 / -0

Let $$P\left ( n \right )= n^{3}-n$$, the largest number by which $$P\left ( n \right )$$ is divisible $$\forall $$ possible integral values of $$n$$ is

A
$$2$$

B
$$3$$

C
$$5$$

D
$$6$$

Solution

$$P\left( n \right) =n^{ 3 }-n=\left( n-1 \right) n\left( n+1 \right) $$
Therefore, $$P(n)$$ is the product of $$3$$ consecutive natural numbers
which is divisible by $$3!=6$$

Ans: D
Question 10
1 / -0

For every natural number n -

A
$$n\, >\, 2^{n}$$

B
$$n\, <\, 2^{n}$$

C
$$n\, / 2^{n}$$

D
$$n\, / 2^{2n}$$

Solution

Consider a function
$$f(n)=2^{n}-n$$

Therefore
$$f'(n)=2^{n}log2-1$$

Now
$$f'(n)>0$$ implies

$$2^{n}log(2)>1$$

$$2^{n}>\dfrac{loge}{log2}$$

$$2^{n}>log_{2}(e)$$

$$nlog2>log(log_{2}(e))$$

$$n>log_{2}(log_{2}(e))$$

Now
$$2<e<3$$
$$1<log_{2}(e)<log_{2}(3)$$

$$1<log_{2}(e)<1.58$$

$$log_{2}(1)<log_{2}(e)<log_{2}(1.58)$$

$$0<log_{2}(log_{2}(e))<0.7$$
Hence
$$n>log_{2}(log_{2}(e))$$

$$n>0.7$$
Thus
$$f(n)$$ is increasing for all $$n>0.7$$.
But $$n$$ is a natural number.
Hence
$$f(n)$$ is increasing for all natural numbers $$n\in N$$
Hence
$$2^{n}-n>0 $$
$$2^{n}>n$$ for all $$n\in N$$

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Principle of Mathematical Induction Test - 14

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Principle of Mathematical Induction Test - 14

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SHARING IS CARING

Question 1

$$2$$

$$3$$

$$5$$

none of these

Solution

Question 2

$$P\left ( 1 \right )$$

$$P\left ( 2 \right )$$

$$P\left ( 3 \right )$$

None of these.

Solution

Question 3

only $$1$$

$$1$$ and $$8$$

only $$8$$

$$1, 8, 49$$

Solution

Question 4

$$P\left ( 1 \right )$$

$$P\left ( 2 \right )$$

$$P\left ( 3 \right )$$

$$P\left ( k+1 \right )$$

Solution

Question 5

$$11$$

$$9$$

$$99$$

none of these

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 7

$$2$$

$$3$$

$$5$$

$$6$$

Solution

Question 8

only for $$n> 1$$.

only for $$n$$ is an odd positive integer.

only for $$n$$ is an even positive integer.

$$n\in N$$

Solution

Question 9

$$2$$

$$3$$

$$5$$

$$6$$

Solution

Question 10

$$n\, >\, 2^{n}$$

$$n\, <\, 2^{n}$$

$$n\, / 2^{n}$$

$$n\, / 2^{2n}$$

Solution

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