MCQ Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 3

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

0%
$11$
0%
$9$
0%
$99$
0%
none of these

Explanation

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.

For all

$n\in I^+$ , the statement

$P\left ( n \right )= \displaystyle \frac{n^{7}}{7}+\displaystyle \frac{n^{5}}{5}+\frac{2}{3}n^{3}-\frac{n}{105}$ is a natural number is true, if

0%
$P\left ( 1 \right )$ is true
0%
$P\left ( 2 \right )$ is true
0%
$P\left ( 3 \right )$ is true
0%
True $\forall \;n\;\in\;N$

Explanation

Given

$P\left ( n \right )= \displaystyle \frac{n^{7}}{7}+\displaystyle \frac{n^{5}}{5}+\frac{2}{3}n^{3}-\frac{n}{105}$

Put

$n=1$ in the given expression

$\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{1}{105}$

$=\dfrac{105}{105}=1$
which is a natural number.
Hence, P(1) is true.

Put

$n=2$ in the given expression

$\dfrac{2^7}{7}+\dfrac{2^5}{5}+\dfrac{2^4}{3}-\dfrac{2}{105}$

$=\dfrac{3150}{105}=30$
which is a natural number.
Hence, P(2) is true.

Given that

$P(n)$ is true

$\displaystyle \frac{n^{7}}{7}+\displaystyle \frac{n^{5}}{5}+\frac{2}{3}n^{3}-\frac{n}{105}$ is a natural number
So, we will check for

$P(n+1)$
Consider,

$\displaystyle \frac { (n+1)^{ 7 } }{ 7 } +\frac { (n+1)^{ 5 } }{ 5 } +\frac { 2 }{ 3 } (n+1)^{ 3 }-\frac { n+1 }{ 105 }$

$=\displaystyle \frac { n^{ 7 }+7n^{ 6 }+21n^{ 5 }+35n^{ 4 }+35n^{ 3 }+21n^{ 2 }+7n+1 }{ 7 } +\frac { n^{ 5 }+5n^{ 4 }+10n^{ 3 }+10n^{ 2 }+5n+1 }{ 5 } +\frac { 2 }{ 3 } (n^{ 3 }+3n^{ 2 }+n+1)-\frac { n+1 }{ 105 }$

$=\displaystyle (\frac { (n+1)^{ 7 } }{ 7 } +\frac { (n+1)^{ 5 } }{ 5 } +\frac { 2 }{ 3 } (n+1)^{ 3 }-\frac { n+1 }{ 105 } )+(n^6+3n^5+6n^4+7n^3+7n^2+4n)+(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{1}{105})$

$=P(n)+\text{natural number}+P(1)$
So,

$P(n+1)$ is a natural number.
Hence, by mathematical induction method , the given result is true for all

$n\in I^+$

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

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

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.

$n\in N$ , then

$n^{3}+2n$ is divisible by

0%
$2$
0%
$3$
0%
$5$
0%
$6$

Explanation

$f\left( n \right) =n^{ 3 }+2n$
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

$\displaystyle P\left ( n \right )=\frac{n^{5}}{5}+\frac{n^{3}}{3}+\frac{7n}{15}$ , then statement "

$P(n)$ is a natural number" is true,

0%
only for $n> 1$ .
0%
only for $n$ is an odd positive integer.
0%
only for $n$ is an even positive integer.
0%
$n\in N$

Explanation

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

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

0%
$2$
0%
$3$
0%
$5$
0%
$6$

Explanation

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

$P(n)$ is the product of

$3$ consecutive natural numbers
which is divisible by

$3!=6$

Ans: D

For every natural number n -

0%
$n\, >\, 2^{n}$
0%
$n\, <\, 2^{n}$
0%
$n\, / 2^{n}$
0%
$n\, / 2^{2n}$

Explanation

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

$n\in N$

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

$n$ radical signs, then by method of mathematical induction which of the following is true?

0%
$\displaystyle a_{n}>6,\forall n> 1$
0%
$\displaystyle a_{n}>3,\forall n> 1$
0%
$\displaystyle a_{n}>4,\forall n> 1$
0%
$\displaystyle a_{n}<2,\forall n> 1$

Explanation

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

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

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

But

$a_n>0 \displaystyle \therefore a_{n}=\frac{1+\sqrt{29}}{2} >3$

0%
Both (A) & (R) are individually true & (R) is correct explanation of (A).
0%
Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
0%
(A) is true but (R) is false.
0%
(A) is false but (R) is true.

Explanation

$\quad\dfrac { n }{ 2 } >\dfrac { n }{ 3 } \\ \Rightarrow { \left( \dfrac { n }{ 2 } \right) }^{ n }>{ \left( \dfrac { n }{ 3 } \right) }^{ n }$

Also,

$\dfrac { n }{ 2 } \times \dfrac { n }{ 2 } \times$ ...

$\times \dfrac { n }{ 2 } \times \dfrac { n }{ 2 }>n\times \left( n-1 \right) \times \left( n-2 \right) \times$ ...

$\times 2\times 1$

$\Rightarrow { \left( \dfrac { n }{ 2 } \right) }^{ n }>{ n! }$

Hence, option

$A$ is correct.

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 -

0%
n > 1
0%
n
0%
n > 2
0%
None of these

Explanation

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

The inequality

$n!\, >\, 2^{n\, -\, 1}$ is true -

0%
For all n > 1
0%
For all n > 2
0%
For all $n\, \epsilon\, N$
0%
None of these

Explanation

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

$3!>{ 2 }^{ 3-1 }$ 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

$D$

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

0%
$P\left ( n \right )$ s divisible by $x+y$ only for odd positive integer
0%
$P\left ( n \right )$ is divisible by $x+y$ only for positive even integer
0%
$P\left ( n \right )$ is true $\forall n\in N$
0%
$P\left ( k \right )$ is true

Explanation

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

$\forall n\in N$ .

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

0%
Negative integers
0%
Positive integers
0%
Rational numbers
0%
None of these

Explanation

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.

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

0%
$19$
0%
$7$
0%
$133$
0%
none of these

Explanation

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

For positive integer n,

$3^{n} < n!$ when

0%
$n \geq 6$
0%
$n > 7$
0%
$n \geq 7$
0%
$n \leq 7$

Explanation

$3^n<n!$

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

$n\ge7$ .

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

0%
n > 1
0%
n
0%
n > 2
0%
none of these

Explanation

$P(n)=n^{2}+n$

$=n(n+1)$
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.

For natural number n,

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

0%
$n < 2$
0%
$n > 2$
0%
$n \geq 2$
0%
Never

Explanation

${ 2 }^{ n }\left( n-1 \right) !<n^{ n }$

$n=1$

Then,

${ 2 }^{ 1 }\left( 1-1 \right) !<1$

$\Rightarrow { 2 }\times 0!<1$

$\Rightarrow { 2 }\times 1<1$ which is not possible

$n=2$

Then,

${ 2 }^{ 2 }\left( 2-1 \right) !<{ 2 }^{ 2 }$

$\Rightarrow { 4 }\times 1!<4$ which is not possible.

$n=3$ 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

$n>2$

For every positive integer
n,

$\displaystyle \frac{n^{7}}{7} + \frac{n^{5}}{5} + \frac {2n^{3}}{3} - \frac{n}{105}$ is

0%
an integer
0%
a rational number
0%
a negative real number
0%
an odd integer

Explanation

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.

$P$ is a prime number then

$n^{p} - n$ is divisible by

$p$ when

$n$ is a

0%
natural number greater than 1
0%
odd number
0%
even number
0%
None of these

Explanation

Let

$P(n) :$

$n^{p} - n$

when

$p = 2$

$P(n) =$

$n^{2} - n$

$P(1) = 0$ which is divisible all n

$\in$ N

$P(2) = 2$ which is divisible by

$2$

$P(3) = 6$ which is divisible by

$2$

Hence

$P(n)$ is divisible by

$2$ when

$n$ is greater than

$1$

The difference between an

$+$ ve integer and its cube, is divisible by

0%
$4$
0%
$6$
0%
$9$
0%
None of these

Explanation

Let

$n$ be the positive integer.

Now its cube is

${ n }^{ 3 }$

We have to check for divisibility of

${ n }^{ 3 }-n$

$\Rightarrow { n }^{ 3 }-n=n\left( n-1 \right) \left( n+1 \right) =\left( n-1 \right) n\left( n+1 \right)$

This clearly is the product of

$3$ consecutive numbers.

The product of

$3$ consecutive numbers is divisible by

$6$ .

As per the divisibility rule for

$6,$ the number should be divisible both by

$2$ and

$3.$

In the product of

$3$ consecutive numbers, the number will turn out to be

divisible by both

$2$ and

$3.$

This can be proved by Mathematical Induction.

If n is a natural number then

$\left( \displaystyle \frac{n + 1}{2} \right)^{n}\, \geq n!$ is true when.

0%
n > 1
0%
$n \geq 1$
0%
n > 2
0%
Never

Explanation

$n$ is a natural number and

$\left(\dfrac{n+1}{2}\right)^n$

Verifying for

$n=1$

$\left(\dfrac{1+1}{2}\right)^1= \left(\dfrac{2}{2}\right)=1\ge{1}!$

Therefore, the condition is satisfied.

From option

$\left(\dfrac{n+1}{2}\right)^{n}\ge{n}!$

For

$n\ge1$ and also

$n=0$

Hence, option B is the correct answer.

For every natural number

$n$ ,

$n(n + 3)$ is always :

0%
multiple of $4$
0%
multiple of $5$
0%
even
0%
odd

Explanation

$n(n+3)=n^2+3n+2-2=(n+1)(n+2)-2\equiv\text{even}-\text{even}\equiv \text{even}$

Since for any natural number

$n$ ,

$(n+1)(n+2)$ will always be an even number.

We know there are two types of natural numbers even and odd

Case 1. If

$n$ is even then

$n+3$ will be odd

Thus

$n(n+3) =$ even

$\times$ odd

$=$ even

Case 2. If

$n$ is odd then

$n+3$ will be even

Thus

$n(n+3) =$ odd

$\times$ even

$=$ even

Hence

$n(n+3)$ will always be even number

For all n

$\in$ N,

$n^{4}$ is less than

0%
$10^{n}$
0%
$4^{n}$
0%
$10^{10}$
0%
None of these

Explanation

For

$n=1$ ,

$n^4=1$

Option A : 10

Option B: 4

For

$n=2$ ,

$n^4=16$

Option A: 100

Option B: 16

Thus,

$n$ is always less than

$10^n$ .

A student was asked to prove a statement by induction. He proved
(i) P(5) is true and
(ii) Trutyh of P(n)

$\Rightarrow$ truth of p(n + 1), n

$\in$ N
On the basis of this, he could conclude that P(n) is true for

0%
no n $\in$ N
0%
all n $\in$ N
0%
all n $\geq$ 5
0%
None of these

Explanation

$P(a)$ is true and truth of

$P(n)$ implies that

$P(n+1)$ is also true, then, we can conclude that

$P(n)$ is true for all

$n\geq a$ where

$n\in N$ .
Here

$P(5)$ is true.
Hence we can conclude that

$P(n)$ is true for all

$n\geq 5$

For every positive integral value of n,

$3^n > n^3$ when

0%
n > 2
0%
$n\geq 3$
0%
$n\geq 4$
0%
n < 4

Explanation

Let

$P(n) : 3^n > n^3$

$P(1) : 3 > 1$ , which is true

$P(2) : 3^2 > 2^3$ , which is true

$P(3) : 3^3 > 3^3$ , which is false

$P(4) : 3^4 > 4^3$ , which is true

$P(5) : 3^5 > 5^3$ , which is true etc.

Hence

$P(n)$ is true when

$n > 4.$

For every positive integer

$n, \dfrac {n^7}{7}+\dfrac {n^5}{5}+\dfrac {2n^3}{3}-\dfrac {n}{105}$ is

0%
an integer
0%
a rational number
0%
a negative real number
0%
an odd integer

Explanation

$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 }$

where

$n$ is a positive integer.

When

$n=1$ , we get,

$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 }$

$=\dfrac { 15{ n }^{ 7 }+21{ n }^{ 5 }+70{ n }^{ 3 }-n }{ 105 }$

$=\dfrac { 15+21+70-1 }{ 105 } =\dfrac { 105 }{ 105 } =1$ which is an integer.

When

$n=2$ ,

we get;

$\dfrac { { n }^{ 7 } }{ 7 } +\dfrac { { n }^{ 5 } }{ 5 } +\dfrac { 2{ n }^{ 3 } }{ 3 } -\dfrac { n }{ 105 }$

$=\dfrac { 15{ n }^{ 7 }+21{ n }^{ 5 }+70{ n }^{ 3 }-n }{ 105 }$

$=\dfrac { \left( 15\times 128 \right) +\left( 21\times 32 \right) +\left( 70\times 28 \right) -2 }{ 105 }$

$=\dfrac { 1920+672+560-2 }{ 105 }$

$=\dfrac { 3105 }{ 105 } =30$ which is an even positive integer.

$P(n) : 3^{2n+2} -8n -9$ is divisible by 64, is true for

0%
all $n\epsilon N \cup \left \{0\right \}$
0%
$n\geq 2, n\epsilon N$
0%
$n\epsilon N, n > 2$
0%
none of these

The smallest positive integer for which the statement

$3^{n+1} < 4^n$ holds is

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Let

$P(n) : 3^{n + 1} < 4^n$

$P(1) : 3^2 < 4$ which is false

$P(2) : 3^3 < 4^2$ which is false

$P(3) : 3^4 < 4^3$ which is false

$P(4) : 3^5 < 4^4$ which is true

If x > 1, then the statement

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

0%
all $n\epsilon N$
0%
all n > 1
0%
all n > 1 and $x\neq 0$
0%
None of these

Explanation

$x > 1$ 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

$P(n)$ is true for all

$n > 1$ and

$x\neq 0$

If n is a natural number then

$\left (\frac {n+1}{2}\right )^n \geq n!$ is true when

0%
n > 1
0%
$n\geq 1$
0%
n > 2
0%
never

Explanation

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.

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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