Principle of Mathematical Induction Test

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Principle of Mathematical Induction Test - 15

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Question 1
1 / -0

If $$\displaystyle a_{n}=\sqrt{7+\sqrt{7+\sqrt{7+.....}}} $$ having $$n$$ radical signs, then by method of mathematical induction which of the following is true?

A
$$\displaystyle a_{n}>6,\forall n> 1$$

B
$$\displaystyle a_{n}>3,\forall n> 1$$

C
$$\displaystyle a_{n}>4,\forall n> 1$$

D
$$\displaystyle a_{n}<2,\forall n> 1$$

Solution

$$\displaystyle a_{n}=\sqrt{7+\sqrt{7+\sqrt7+.....}} $$

$$\Rightarrow \displaystyle

a_{n}=\sqrt{7+a_{n}}\Rightarrow a_{n}^{2}-a_{n}-7=0$$

$$\displaystyle \therefore

a_{n}=\frac{1\pm\sqrt{1+28}}{2}=\frac{1\pm\sqrt{29}}{2}$$

But $$a_n>0 \displaystyle \therefore

a_{n}=\frac{1+\sqrt{29}}{2} >3$$
Question 2
1 / -0

Let $$P(n)\, :\, n^{2}\, +\, n$$ is an odd integer. It is seen that truth of $$P(n)\, \Rightarrow$$ the truth of P(n + 1). Therefore, P(n) is true for all -

A
n > 1

B
n

C
n > 2

D
None of these

Solution

Since $$P(1): {1}^{2}+1=2$$ is not an odd integer, so $$P(1)$$ is not true.
$$\therefore$$ Principal of induetion is not applicable.
In fact, $$p(n)=n(n+1)$$ being the product of two conseative integers is always even
Question 3
1 / -0

Let $$P\left ( n \right )= x^{2n-1}+y^{2n-1}$$ is divisible by $$x+y$$ as $$P\left ( 1 \right )$$ is true, then truth of $$P\left ( k+1 \right )$$ indicates

A
$$P\left ( n \right )$$ s divisible by $$x+y$$ only for odd positive integer

B
$$P\left ( n \right )$$ is divisible by $$x+y$$ only for positive even integer

C
$$P\left ( n \right )$$ is true $$\forall n\in N$$

D
$$P\left ( k \right )$$ is true

Solution

This is the basic principle of mathematic induction that if P(1) is true and P(k+1) is true, then P(n) must be true $$\forall n\in N$$.
Question 4
1 / -0

The sequence $$\displaystyle \left ( x_{n}n\geq 1 \right )$$ is defined by $$\displaystyle x_{1}=0 $$ and $$\displaystyle x_{n+1}=5x_{n}+\sqrt{24x^{2}_{n}+1} $$ for all $$\displaystyle n\geq 1.$$ Then all $$\displaystyle x_{n} $$ are

A
Negative integers

B
Positive integers

C
Rational numbers

D
None of these

Solution

We first notice that the sequence is increasing and all of its terms are positive. Next we observe that the recursive relation is equivalent to $$\displaystyle x^{2}_{n+1}-10x_{n}x_{n+1}+x^{2}_{n}-1=0$$ Replacing $$n$$ by $$n-1$$ yields $$\displaystyle x^{2}_{n}-10x_{n}x_{n-1}+x^{2}_{n-1}-1=0,$$ hence for$$\displaystyle n\geq 2$$ the numbers $$\displaystyle x_{n+1}$$ and $$\displaystyle x_{n-1}$$ are distinct roots of the equation $$\displaystyle x^{2}-10xx_{n}+x^{2}_{n}-1=0.$$ The Viete's relations yield $$\displaystyle x_{n+1}+x_{n-1}=10x_{n},$$ or, $$\displaystyle x_{n+1}=10x_{n}-x_{n-1}$$ for all $$\displaystyle n\geq 2$$ . Because $$x_{ 1 }=1$$ and $$x_{ 2 }=10$$ it follows inductively that all $$x_n$$ are positive integers.
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

This question can be solved by method of induction
Assume $$n!>{ 2 }^{ n-1 }$$ to prove $$n+1!>{ 2 }^{ n}$$
so we need to find the lowest natural number which satisfies our assumption that is 3
as $$3!>{ 2 }^{ 3-1 }$$ as $$6>4$$
hence n>2 and n natural number now we need to solve it by induction
to prove $$n+1!>{ 2 }^{ n}$$
we know $$n!>{ 2 }^{ n-1 }$$
multiplying n+1 on both sides we get $$n+1!>{ 2 }^{ n-1 }\left( n+1 \right) $$
$$n>2$$ hence $$n+1>3$$
which also implies $$n+1>2$$
hence $$n+1!>{ 2 }^{ n}$$
hence $$n!>{ 2 }^{ n-1 }$$ proved by induction where $$n\in N$$ and n>2
n is a natural number is missing in the option
hence $$D$$
Question 6
1 / -0

Let $$P\left ( n \right )=11^{n+2}+12^{2n+1}$$, then the least value of the following which $$P\left ( n \right )$$ is divisible by is

A
$$19$$

B
$$7$$

C
$$133$$

D
none of these

Solution

Given that $$P\left( n \right) =11^{ n+2 }+12^{ 2n+1 }$$
Put $$n=1$$ to obtain $$P\left( 1 \right) =11^{ 1+2 }+12^{ 2+1 }=3059=7\left( 437 \right) $$
Therefore, $$P\left( 1 \right) $$ is divisible by $$7$$
Assume that for $$n=k$$, $$P\left( k \right) =11^{ k+2 }+12^{ 2k+1 }$$ is divisible by $$7$$
Now, $$P\left( k+1 \right) =11^{ k+3 }+12^{ 2k+3 }=11.11^{ k+2 }+144.12^{ 2k+1 }=11\left( 11^{ k+2 }+12^{ 2k+1 } \right) +133$$
$$\Rightarrow P\left( k+1 \right) =11P\left( k \right) +7\left( 79 \right) $$
SInce, $$P\left( k \right) $$ is divisible by $$7$$
Therefore, $$P\left( k+1 \right) $$ is divisible by $$7$$
And from the principle of mathematical induction $$P\left( n \right) $$ is divisible by $$7$$ for all $$n\in N$$

Ans: B
Question 7
1 / -0

For positive integer n, $$3^{n} < n!$$ when

A
$$n \geq 6$$

B
$$n > 7$$

C
$$n \geq 7$$

D
$$n \leq 7$$

Solution

$$3^n<n!$$

From options:

For $$n=6$$,

$$LHS=729$$ ; $$RHS=720$$

$$\Rightarrow{LHS}>RHS$$

For $$n=7$$,

$$LHS=2187$$ ; $$RHS=5040$$

$$\Rightarrow {LHS}<RHS$$

So, the given condition is true for $$n\ge7$$.
Question 8
1 / -0

Let $$P(n) : n^2 + n$$ is an odd integer. It is seen that truth of $$P(n)\Rightarrow$$ the truth of P(n + 1). Therefore, P(n) is true for all

A
n > 1

B
n

C
n > 2

D
none of these

Solution

$$P(n)=n^{2}+n$$
$$=n(n+1)$$
Hence $$P(n)$$ represents product of two consecutive Natural numbers.
We know that every odd natural number is succeeded by an even natural number. Hence product of two natural numbers will always be even.
Hence
$$P(n)$$ is even for all $$n\in N$$.
Hence the conclusion that $$P(n)=n^{2}+n$$ is an odd integer is false.
Question 9
1 / -0

For natural number n, $$2^{n}\, (n - 1) ! <n^{n}$$, if.

A
$$n < 2$$

B
$$n > 2$$

C
$$n \geq 2$$

D
Never

Solution

$${ 2 }^{ n }\left( n-1 \right) !<n^{ n }$$
If $$n=1$$
Then, $${ 2 }^{ 1 }\left( 1-1 \right) !<1$$
$$\Rightarrow { 2 }\times 0!<1$$
$$\Rightarrow { 2 }\times 1<1$$ which is not possible
If $$n=2$$
Then, $${ 2 }^{ 2 }\left( 2-1 \right) !<{ 2 }^{ 2 }$$
$$\Rightarrow { 4 }\times 1!<4$$ which is not possible.
If $$n=3$$ then,
$${ 2 }^{ n }\left( n-1 \right) !<n^{ n }$$
$${ 2 }^{ 3 }\left( 3-1 \right) !<{ 3 }^{ 3 }$$
$$\Rightarrow { 4}\times 2!<27$$
$$16<27$$ which is possible.
Therefore, for natural number n, $${ 2 }^{ n }\left( n-1 \right) !<n^{ n }$$ if $$n>2$$
Question 10
1 / -0

For every positive integer
n, $$\displaystyle \frac{n^{7}}{7} + \frac{n^{5}}{5} + \frac {2n^{3}}{3} - \frac{n}{105}$$ is

A
an integer

B
a rational number

C
a negative real number

D
an odd integer

Solution

Let P(n) =$$\displaystyle \frac{n^{7}}{7} + \frac{n^{5}}{5} + {2n^{3}}{3} - \frac{n}{105}$$
P(1) = $$\displaystyle \frac{1}{7} + \frac {1}{5} + \frac{2}{3} - \frac {1}{105}$$ = 1 (integer)
P(2) = $$2^{4} \displaystyle \left(\frac{8}{7} + \frac{2}{5} + \frac{1}{3} \right) - \frac{2}{105}$$ = 15 (integer) etc.
Hence P(n) is an integer.

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Principle of Mathematical Induction Test - 15

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Principle of Mathematical Induction Test - 15

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Question 1

$$\displaystyle a_{n}>6,\forall n> 1$$

$$\displaystyle a_{n}>3,\forall n> 1$$

$$\displaystyle a_{n}>4,\forall n> 1$$

$$\displaystyle a_{n}<2,\forall n> 1$$

Solution

Question 2

n > 1

n

n > 2

None of these

Solution

Question 3

$$P\left ( n \right )$$ s divisible by $$x+y$$ only for odd positive integer

$$P\left ( n \right )$$ is divisible by $$x+y$$ only for positive even integer

$$P\left ( n \right )$$ is true $$\forall n\in N$$

$$P\left ( k \right )$$ is true

Solution

Question 4

Negative integers

Positive integers

Rational numbers

None of these

Solution

Question 5

For all n > 1

For all n > 2

For all $$n\, \epsilon\, N$$

None of these

Solution

Question 6

$$19$$

$$7$$

$$133$$

none of these

Solution

Question 7

$$n \geq 6$$

$$n > 7$$

$$n \geq 7$$

$$n \leq 7$$

Solution

Question 8

n > 1

n

n > 2

none of these

Solution

Question 9

$$n < 2$$

$$n > 2$$

$$n \geq 2$$

Never

Solution

Question 10

an integer

a rational number

a negative real number

an odd integer

Solution

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