MCQ Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 2

$\forall n\in N; 10^{2n - 1}+1$ is divisible by

0%
$2$
0%
$3$
0%
$7$
0%
$11$

Explanation

For

$n = 1$ , we have

$10^{2n - 1} + 1 = 10 + 1 = 11$ E, which is divisible by

$11$ .

For

$n = 2$ , we have

$10^{2n - 1} + 1 = 10^3 + 1 = 1001$ , which is divisible by

$11$ .

$\forall n\in N; 10^{2n - 1}+1$ is divisible by 11.

Hence, option (d) is correct.

$\forall n\in N; 3n^{5} + 5n^3 + 7n\>$ is divisible by

0%
$3$
0%
$5$
0%
$10$
0%
$15$

Explanation

Put

$n=1$ in the given expression

$3n^{5} + 5n^3 + 7n$

$3+5+7=15$
which is divisible by

$3,5, 15$
Put

$n=2$

$3(32)+5(8)+14=150$
which is divisible by

$3,5,10,15$
Since, P(1) and P(2) both are divisible by

$3,5, 15$
Every number divisible by 15 is divisible by 3 and 5 also.
So, let us assume

$P(n)=3n^{5} + 5n^3 + 7n$ is divisible by 15

$\Rightarrow 3n^{5} + 5n^3 + 7n=15\lambda$
Now, we have to prove P(n+1) is true
Consider,

$P(n+1)=3(n+1)^{5} + 5(n+1)^3 + 7(n+1)$

$=3(n^5+5n^4+10n^3+10n^2+5n+1)+5(n^3+1+3n^2+3n)+7n+7$

$=(3n^5+5n^3+7n)+15n^4+30n^3+45n^2+30n+15$

$=15\lambda +15(n^4+2n^3+3n^2+2n+1)$

$=15 (\lambda+n^4+2n^3+3n^2+2n+1)$
which is divisible by

$15.$
Hence,

$P(n+1)$ is divisible by

$15.$
Hence, by mathematical induction, given expression is divisible by 15, for all

$n\in N$
Hence, given expression is divisible by

$3,5, 15$

$x^n - 1$ is divisible by

$x - k$ , then the least positive integral value of

$k$ is

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

Explanation

$f(x)=x^n-1$ is divisible by

$x-k$

Then

$f(k)=0$

Therefore,

$k^n=1$

and thus least positive integral value of k is 1

$\forall n\in N, n^4$ is less than

0%
$10^n$
0%
$4n$
0%
$4^n$
0%
$10^{10}$

Explanation

Let

$y =n^4$
put

$n=10\Rightarrow y = 10^4=10000>4\times 10, 4^4$
Hence option B and C discarded.
Now put

$n = 10^3\Rightarrow y = 10^{12}>10^{10}$ . thus option D is also incorrect.
Hence option 'A' is correct.

$n$

$\in$ N, then

$x^{2n - 1} + y^{2n - 1}$ is divisible by

0%
$x + y$
0%
$x - y$
0%
$x^2 + y^2$
0%
none of these

Explanation

Le4 the given statement be

$P\left( n \right)$

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

We check

$P\left( n \right)$ for

$n=1$

$P\left( 1 \right) ={ x }^{ 2-1 }+{ y }^{ 2-1 }=x+y$

Thus

$P\left( 1 \right)$ is divisible by

$x+y$

Let us assume that

$P\left( n \right)$ is divisible by

$x+y$ when

$n=k$

Now for

$n=k+1$ we have

$P\left( k+1 \right) ={ x }^{ 2k+1 }+{ y }^{ 2k+1 }$

$\Rightarrow P\left( k+1 \right) =\left( { x }^{ 2 } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 } \right)$

$\Rightarrow P\left( k+1 \right) =\left( { x }^{ 2 } \right) \left( { x }^{ 2k-1 }-{ y }^{ 2 } \right) \left( { x }^{ 2k-1 }+{ y }^{ 2 } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 } \right)$

$\Rightarrow P\left( k+1 \right) =\left( { x }^{ 2 }-{ y }^{ 2 } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 }+{ x }^{ 2k-1 } \right)$

$\Rightarrow P\left( k+1 \right) =\left( { x }-{ y } \right) \left( { x }+{ y } \right) \left( { x }^{ 2k-1 } \right) +\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 }+{ x }^{ 2k-1 } \right)$

Now

$\left( { x }-{ y } \right) \left( { x }+{ y } \right) \left( { x }^{ 2k-1 } \right)$ and

$\left( { y }^{ 2 } \right) \left( { y }^{ 2k-1 }+{ x }^{ 2k-1 } \right)$ are divisible by

$x+y$

${ x }^{ 2k+1 }+{ y }^{ 2k+1 }$ is also divisible by

$x+y$

${ x }^{ 2n-1 }+{ y }^{ 2n-1 }$ is divisible by

$x+y$ when

$n=k+1$ if it is divisible by

$x+y$ when

$n=k$

so by principle of mathematical induction,

${ x }^{ 2n-1 }+{ y }^{ 2n-1 }$ is divisible by

$x+y$

For all

$n\in N, \sum n$ is

0%
$\displaystyle < \frac{(2n + 1)^2}{8}$
0%
$\displaystyle > \frac{(2n + 1)^2}{8}$
0%
$\displaystyle = \frac{(2n + 1)^2}{8}$
0%
none of these

Explanation

$\displaystyle \sum n = \dfrac{n(n+1)}2=\dfrac{n^2+n}2=\dfrac{4n^2+4n}{8}$

$\quad = \dfrac{4n^2+4n+1-1}{8}=\dfrac{(2n+1)^2-1}{8}$

$\quad =\dfrac{(2n+1)^2}{8}-\dfrac{1}{8}<\dfrac{(2n+1)^2}{8}$

Hence option 'A' is correct choice.

$49^{n} + 16 n + \lambda$ is divisible by

$64$ for all

$n\in N$ , then the least negative integral value of

$\lambda$ is

0%
$-12$
0%
$-1$
0%
$-3$
0%
$-4$

Explanation

For

$n = 1$ , we have

$49^n + 16 n + \lambda = 49 + 16 + \lambda = 65 + \lambda= 64 + (\lambda + 1)$ , which is divisible by 64 if

$\lambda = -1$ .
For

$n = 2$ , we have

$49^n + 16 n + \lambda = 49^2 + 16 \times 2 + \lambda = 2433 + \lambda= (64 \times 38) + (\lambda + 1)$ , which is divisible by 64 if

$\lambda = -1$
Hence,

$\lambda = -1$ .

Let

$P(m)$ be the statement

$m^{2}> 100$ , the statement

$P(k + 1)$ will be true if

0%
$P(1)$ is true
0%
$P(2)$ is true
0%
$P(k)$ is true
0%
none of these

Explanation

$P(r)$ is true

$\displaystyle \Rightarrow r^{2}> 100$

$\displaystyle \Rightarrow r^{2}+2r+1> 100+2r+1$

$\displaystyle \Rightarrow \left ( r+1 \right )^{2}> 100$

$\displaystyle \Rightarrow P(r+1)$ is true as

$\displaystyle r^{2}+\left ( 2r+1 \right )>r^{2}> 100$

$\displaystyle \Rightarrow P\left (k+1 \right )$ is true (say

$r=k$ )

$P(k+1)$ is true when every

$p(k)$ is

So, In order to prove that

$P(k+1)$ is true.

It is sufficient to consider

$P(k)$ is true.

The product of three consecutive natural numbers is divisible by

0%
$3$
0%
$8$
0%
$6$
0%
$11$

Explanation

Let

$n, n + 1, n + 2$ be three consecutive natural numbers and P (n) be their product. Then,

$P(n) = n(n + 1) (n+2)$

We have,

$P (1) = 1 \times 2 \times 3 = 6$ , which is divisible by 3 and 6.

$P(2) = 2 \times 3 \times 4 = 24$ , which is divisible by 3, 8 and 6.

$P(3) = 3 \times 4 \times 5 = 60$ , which is divisible by 3 and 6.

$P(4) = 4 \times 5 \times 6 = 120$ , which is divisible by 3, 8 and 6.

Hence, P(n) is divisible by 3 and 6 for all

$n \in$ N.

Let

$P(n)$ be a statement such that truth of

$P\left ( n \right )\Rightarrow$ the truth of

$P\left ( n+1 \right )$ for all

$n\epsilon N$ , then

$P(n)$ is true

0%
$\forall n> 1$
0%
$\forall n$
0%
nothing can be said
0%
$\forall n> k$ (k is some fixed positive integer)

Explanation

Let

$P\left( n \right)$ be a statement such that truth of

$P\left( n \right) \Rightarrow P\left( n+1 \right)$ for all

$n\in N$

Now from the statement of the Principle of Mathematical Induction,

we know that if a statement is assumed true for

$n=k$ and then if it holds true for

$n=k+1$ and

$n=1$ , then the statement is true for all

$n$ .

$P\left( n \right) \Rightarrow P\left( n+1 \right)$ is given so for

$P\left( n \right)$ to be true for all

$n\in N$ ,

$P\left( 1 \right)$ should be true.

But nothing has been stated about

$P\left( 1 \right)$ . so nothing can be said about the truth of

$P\left( n \right)$ .

$\displaystyle x^{3^{n}}+y^{3^{n}}$ is divisible by

$x+y$ , if

0%
$n$ is any integer $\geq0$
0%
$n$ is an odd positive integer
0%
$n$ is an even positive integer
0%
$n$ is a rational number

Statement 1 : For each natural number

$n, (n + 1)^7 - n^7 - 1$ is divisible by 7.
Statement 2 : For each natural

$n$ ,

$n^7 - n$ is divisible by 7.

0%
A) Statement - 1 is True, Statement - 2 is True; Statement -2 is a correct explanation for Statement - 1
0%
B) Statement - 1 is True, Statement - 2 is True; Statement -2 is NOT a correct explanation for Statement - 1
0%
C) Statement - 1 is True, Statement-2 is False
0%
D) Statement - 1 is False, Statement-2 is True

Explanation

It can be checked by using the Principle of mathematical induction that statement - 2 is true for all

$n \in N.$

So, statement - 2 is correct.

Now,

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

Here both the terms are divisible by

$7.$

Therefore,

$(n + 1)^7 - n^7 - 1$ is also divisible by

$7$ .

So, statement - 1 is correct and statement-2 is a correct explanation for statement-1.

For

$n \in N, x^{n + 1} + (x + 1)^{2n - 1}$ is divisible by

0%
$x$
0%
$x + 1$
0%
$x^2 + x + 1$
0%
$x^2 - x + 1$

Explanation

For

$n = 1$ , we have

$x^{n + 1} + (x + 1)^{2n - 1}$

$= x^2 + (x + 1) = x^2 + x + 1$ , which is divisible by

$x^2 + x + 1.$
For

$n = 2$ , we have

$x^{n + 1} + (x + 1)^{2n - 1}$

$= x^3 + (x + 1)^3$

$= (2x + 1) (x^2 + x + 1)$ , which is divisible by

$x^2 + x + 1.$
Hence, option (c) is true.

Let

$P\left ( n \right )=n\left ( n+1 \right )$ is an even number, then which of the following satisfy

$P(n)$

0%
$P(3)$
0%
$P(100)$
0%
$P(50)$
0%
All of these

Explanation

Given,

$P(n)=n(n+1)$

$\displaystyle \therefore P(3)=3.4=12(even)$

$\displaystyle P(100)=100\times 101=10100(even)$

$\displaystyle P(50)=50\times 51=2520(even)$

$P(3), P(100),P(50)$ are even numbers.

Short cut Method:

$P(n)=n(n+1)$

Product of two consecutive integer (natural numbers) is always even.

For each natural number, the statement

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

0%
$10$
0%
$6$
0%
$7$
0%
None of these.

Explanation

$P\left( n \right) =2^{ 3n }-1=8^n-1$

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

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

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

Therefore,

$P(n)$ is divisible by

$7$ .

Let

$P\left ( n \right ):n^{2}+n$ is an odd integer

$P\left ( k \right )\Rightarrow P\left ( k+1 \right )$ is true
Then

$P\left ( n \right )$ is true for all

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

Explanation

Given that

$P\left( n \right) :{ n }^{ 2 }+n$ is an odd integer.

$P\left( k \right) \Rightarrow P\left( k+1 \right)$ is true.

For

$P\left( n \right)$ to be true for all

$n$ , it should be true for

$n=1$

$P\left( 1 \right) ={ 1 }^{ 2 }+1=2$ This is not an odd integer, So not true for

$n=1$

Check for

$n=2$ ,

$P\left( 2 \right) ={ 2 }^{ 2 }+2=6$ This is again not odd,

so not true for

$n=2$

For

$n=3$ ,

$P\left( 3 \right) ={ 3 }^{ 2 }+3=9$ , odd so true

For

$n=4$ ,

$P\left( 4 \right) ={ 4 }^{ 2 }+4=20$ not odd, so not true.

So we see from options, it not true for any given option.

$P(n)$ be the statement

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

0%
$P(2)$
0%
$P(3)$
0%
$P(4)$
0%
none of these

Explanation

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

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

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

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

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

Ans: D

Let

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

$P(n)$ is divisible by

0%
$63$
0%
$36$
0%
$49$
0%
$25$

Explanation

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

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

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

Therefore,

$P(n)$ is divisible by

$49$ .

Ans: 49

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

0%
$1$
0%
$3!$
0%
$3$
0%
$9!$

Explanation

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

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

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

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

Which is divisible by

$9 \displaystyle$

$\therefore$ least value of

$n$ is

$1.$

Let

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

0%
$\forall n$
0%
for $n=1$
0%
$n=2$
0%
none of these

Explanation

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.

$P\left ( n \right ):2^{n+2}< 3^{n}$ , is true for

0%
$n\in N$
0%
$n>3, n\in N$
0%
$n>2, \forall n\in N$
0%
none of these.

Explanation

$\displaystyle P\left ( n \right ):2^{n+2}< 3^n$

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 $\displaystyle \forall n> 3,n\epsilon N$

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

0%
$P(1)$ is true
0%
$P(2)$ is true
0%
$P(m)$ is true
0%
none of these

Explanation

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

$LHS=2+4=6$

$RHS=2(3)+2=8$

Let

$P(n)$ be the statement that

$n^{2}-n+41$ is prime, then which of the following is not true ?

0%
$P(2)$
0%
$P(3)$
0%
$P(41)$
0%
none of these

Explanation

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

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

$\forall k\geq 2$ ?

0%
$2^{k}> 5k> 1$
0%
$2^{k+1}> 2k> k+1$
0%
$2^{k}> 2\left ( k+1 \right )> k$
0%
None of these.

Explanation

$\quad{ 2 }^{ k }>k$ ,

$\forall n=k$

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

$B$ .

The inequality,

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

0%
$n\geq 4$
0%
$n> 1$
0%
$n> 2$
0%
$\forall n\in N$

Explanation

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

$\displaystyle \therefore P\left ( 1 \right )= 1!> 2^{1}$ (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: $n \ge 4$

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

0%
$n> 1$
0%
$n$ is odd
0%
$n$ is even
0%
none of these

Explanation

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

$n=1$

$\displaystyle \therefore p\left ( 1 \right )= a+b$ 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

$k \displaystyle \epsilon N.$

Let

$f\left ( n \right ) = 8^{n}-3^{n}$ , if

$n$ is odd natural number then

$f\left ( n \right )$ is divisible by

0%
$2$
0%
$3$
0%
$5$
0%
none of these

Explanation

$f\left ( n \right ) = 8^{n}-3^{n}$
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

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

0%
$P\left ( 1 \right )$
0%
$P\left ( 2 \right )$
0%
$P\left ( 3 \right )$
0%
None of these.

Explanation

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.

$P\left( 1 \right)$

LHS

$={ 1 }^{ 2 }=1$ , RHS

$=\displaystyle\frac { 1\left( 4{ \left( 1 \right) }^{ 2 }-1 \right) }{ 3 } =1$

So True for

$n=1$ .

$P\left( 2 \right)$

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

$n=2$

$P\left( 3 \right)$

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

$P\left ( n \right )= 1+2+3+\dots+n$ is a perfect square,

$N$ is less than

$100$ , then possible values of

$n$ is/are

0%
only $1$
0%
$1$ and $8$
0%
only $8$
0%
$1, 8, 49$

Explanation

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

$P(1) = 1$

$P(8) = 36 =$ square of

$6$

$P(49) = 1225 =$ square of

$35$

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,

0%
$P\left ( 1 \right )$
0%
$P\left ( 2 \right )$
0%
$P\left ( 3 \right )$
0%
$P\left ( k+1 \right )$

Explanation

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

CBSE Questions for Class 11 Engineering Maths Principle Of Mathematical Induction Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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