Principle of Mathematical Induction Test

Result

Principle of Mathematical Induction Test - 9

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Statement-l: For every natural number $n\geq 2,\ \displaystyle \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots\ldots+\frac{1}{\sqrt{n}}>\sqrt{n}$ .
Statement-2: For every natural number $n\geq 2,\ \sqrt{n(n+1)}<n+1$ .

A
Statement-1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1.

B
Statement-1 is true, Statement-2 is false.

C
Statement-1 is false, Statement-2 is true.

D
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Solution

$\displaystyle P\left( n \right) =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +...+\frac { 1 }{ \sqrt { n } }$
$\displaystyle P\left( 2 \right) =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } >\sqrt { 2 }$
Let us assume that $P(k)$
$\displaystyle =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +...+\frac { 1 }{ \sqrt { k } } >\sqrt { k }$ is true
$\displaystyle \therefore P\left( k+1 \right) =\frac { 1 }{ \sqrt { 1 } } +\frac { 1 }{ \sqrt { 2 } } +...+\frac { 1 }{ \sqrt { k } } +\frac { 1 }{ \sqrt { k+1 } } >\sqrt { k+1 }$ has to be true.
$\displaystyle L.H.S.>\sqrt { k } +\frac { 1 }{ \sqrt { k+1 } } =\frac { \sqrt { k\left( k+1 \right)} +1 }{ \sqrt { k+1 } }$
Since $\sqrt { k\left( k+1 \right) } >k\quad \left( \forall k\ge 0 \right)$
$\displaystyle \therefore \frac { \sqrt { k\left( k+1 \right) }+1 }{ \sqrt { k+1 } } >\frac { k+1 }{ \sqrt { k+1 } } =\sqrt { k+1 }$
Let $P\left( n \right) =\sqrt { n\left( n+1 \right) } <\left( k+1 \right)$
Statement-1 is correct.
$P\left( 2 \right) =\sqrt { 2\times 3 } <3$
If $P\left( k \right) =\sqrt { k\left( k+1 \right) } <\left( k+1 \right)$ is true
Now $P\left( k+1 \right) =\sqrt { \left( k+1 \right) \left( k+2 \right)} <k+2$ has to be true
Since $\left( k+1 \right) <k+2$
$\therefore \sqrt { \left( k+1 \right) \left( k+2 \right) } <\left( k+2 \right)$
Hence Statement-2 is not correct explanation of Statement-1.
Question 2
1 / -0

Consider the statement: $"P(n):n^2-n+41$ is prime". Then which one of the following is true?

A
P $(5)$ is false but P $(3)$ is true

B
Both P $(3)$ and P $(5)$ are false

C
P $(3)$ is false but P $(5)$ is true

D
Both P $(3)$ and P $(5)$ are true

Solution

$P(n): n^2-n+41$ is prime

$P(5)=61$ which is prime

$P(3)=47$ which is also prime.
Question 3
1 / -0

Let $S(k) = 1 + 3 + 5 + .... + (2k - 1) = 3 + k^2$ . Then which of the following is true?

A
Principle of mathematical induction can be used to prove the formula

B
$S (k)$ $\Rightarrow$ $S (k + 1)$

C
$S (k)$ ${\nRightarrow}$ $S (k + 1)$

D
$S (1)$ is correct

Solution

Putting $k=1$ , we get L.H.S=1 and R.H.S=4. Hence $A$ and $D$ are incorrect.
Now, $S\left( k+1 \right) =1+3+5+$ ... $+\left( 2k-1 \right) +\left( 2\left( k+1 \right) -1 \right)$
$\Rightarrow S\left( k+1 \right) =S\left( k \right) +\left( 2k+1 \right)$ [Since $S\left( k \right) =1+3+5+$ ... $+\left( 2k-1 \right)$ ]
$\Rightarrow S\left( k+1 \right) =3+{ k }^{ 2 }+\left( 2k+1 \right)$
$\Rightarrow S\left( k+1 \right) =3+{ \left( k+1 \right) }^{ 2 }$
$B$ is correct option.
Question 4
1 / -0

If A = $$\begin{vmatrix}
1 &0 \\
1& 1
\end{vmatrix} $B and I =$ \begin{vmatrix}
1 &0 \\
0& 1
\end{vmatrix} $,then which one of the following holds for all n$ \geq $$ 1, by
the principle of mathematical indunction

A
$A^{n}=nA-(n-1)l$

B
$A^{n}=2^{n-1}A-(n-1)l$

C
$A^{n}=nA+(n-1)l$

D
$A^{n}=2^{n-1}A+(n-1)l$

Solution

$A=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\\ { A }^{ 2 }=\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\\ { A }^{ 3 }=\begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}\\ \therefore { A }^{ n }=\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}$

and $nA=n\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}=\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}$

$\left( n-1 \right) I=\left( n-1 \right) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}\\ \therefore nA-\left( n-1 \right) I=\begin{bmatrix} n & 0 \\ n & n \end{bmatrix}-\begin{bmatrix} n-1 & 0 \\ 0 & n-1 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ n & 1 \end{bmatrix}={ A }^{ n }$
Question 5
1 / -0

Mathematical Induction is the principle containing the set

A
R

B
N

C
Q

D
Z

Solution

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.
Question 6
1 / -0

Let $P(n)$ be a statement and $P(n)=P(n+1) \forall n\in N$ , then $P(n)$ is true for what values of $n$ ?

A
For all $n$

B
For all $n>1$

C
For all $n>m$ , $m$ being a fixed positive integer

D
Nothing can be said

Solution

Given, $P(n) = P(n+1) \forall n\in N$
Substituting $n-1$ in place of $n$ ,
$P(n-1)=P(n)$
Thus if $P(k)$ is true for some $k$ $\in$ $N$ , then it is true for $k-1$ and $k+1$ .
Thus, it is true $\forall k$ $\in$ $N$
Question 7
1 / -0

For every integer $n\geq 1, (3^{2^{n}}-1)$ is always divisible by

A
$2^{n^2}$

B
$2^{n+4}$

C
$2^{n+2}$

D
$2^{n+3}$

Solution

For $n= 1$ , $3^{2^{1}}-1 = 8$ , which is divisible by $2^{n+2}$ .

Let us assume that $3^{2^m} -1$ is divisible by $2^{m+2}$ for some integral value of $m$ .
Let us consider the expression for $m+1$
$3^{2^{m+1}} -1$
$= (3^{2^{m}} -1) \times (3^{2^{m}} +1)$
The first term is divisible $2^{m+2}$ and the second term is also an even number.
Hence, the term is divisible by $2^{m+2}$ .
Hence, by induction we can prove that it is true for all $m$ .
Question 8
1 / -0

Let $P(n)= 5^{n}-2^{n}$ . $P(n)$ is divisible by $3\lambda$ where $\lambda$ and ${n}$ both are odd positive integers, then the least value of $n$ and $\lambda$ will be

A
$13$

B
$11$

C
$1$

D
$5$

Solution

$5^n-2^n$ is divisible by $5-2=3$ always... Putting $n=\lambda =1$ which is the least odd positive integer, this works to be true.
Hence Option C
Question 9
1 / -0

Let $\mathrm{S}(\mathrm{K})=1+3+5+\ldots\ldots..+(2\mathrm{K}-1)=3+\mathrm{K}^{2}$ . Then which of the following is true?

A
$\mathrm{S}(1)$ is correct

B
$\mathrm{S}(\mathrm{K})\Rightarrow \mathrm{S}(\mathrm{K}+1)$

C
$S(\mathrm{K})\neq$ S(K $+$ 1)

D
Principle of mathematical induction can be used to prove the formula

Solution

$S\left( K \right) =1+3+5+...+\left( 2K-1 \right) =3+{ K }^{ 2 }$

Put $K=1$ in both sides

$\therefore L.H.S=1$ and $R.H.S=3+1=4$

$\Rightarrow L.H.S\neq R.H.S$

Put $\left( K+1 \right)$ on both sides in the place of $k$

$L.H.S=1+3+5+...+\left( 2K-1 \right) +\left( 2K+1 \right)$

and $R.H.S=3+{ \left( K+1 \right) }^{ 2 }=3+{ K }^{ 2 }+2K+1$

Let $L.H.S = R.H.S$

$\Rightarrow 1+3+5+...+\left( 2K-1 \right) +\left( 2K+1 \right) =3+{ K }^{ 2 }+2K+1\\ \Rightarrow 1+3+5+...+\left( 2K-1 \right) =3+{ K }^{ 2 }$

If $S(K)$ is true, then $S(K+1)$ is also true.

Hence $S\left( K \right) \Rightarrow S\left( K+1 \right)$
Question 10
1 / -0

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

A
$x-y$

B
$x+y$

C
$xy$

D
$x^{2}+y^{2}$

Solution

$$\displaystyle P\left ( n \right )=x^{2n-1}+y^{2n-1}\forall n \epsilon N.
$$
Substitute $n=1$ to obtain $\displaystyle p\left ( 1 \right )= x+y$ ,Which is divisible by $x+y$ .
For, $n=2$ , we get $\displaystyle P\left ( 2 \right )=x^{3}+y^{3}=\left ( x+y \right )\left ( x^{2}-xy+y^{2} \right )$ which is divisible by $x+y$ .
With the help of induction we conclude that $P(n)$ will be divisible by $x+y$ for all $n\in N$ .

Ans: B

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Principle of Mathematical Induction Test - 9

Result

Principle of Mathematical Induction Test - 9

Score

-

Rank

-

SHARING IS CARING

Question 1

Statement-1 is true, Statement-2 is true; Statement -2 is a correct explanation for Statement-1.

Statement-1 is true, Statement-2 is false.

Statement-1 is false, Statement-2 is true.

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Solution

Question 2

P(5)(5)(5) is false but P(3)(3)(3) is true

Both P(3)(3)(3) and P(5)(5)(5) are false

P(3)(3)(3) is false but P(5)(5)(5) is true

Both P(3)(3)(3) and P(5)(5)(5) are true

Solution

Question 3

Principle of mathematical induction can be used to prove the formula

S(k)S (k)S(k) ⇒\Rightarrow⇒ S(k+1)S (k + 1)S(k+1)

S(k)S (k)S(k) ⇏{\nRightarrow}⇏ S(k+1)S (k + 1)S(k+1)

S(1)S (1)S(1) is correct

Solution

Question 4

An=nA−(n−1)lA^{n}=nA-(n-1)lAn=nA−(n−1)l

An=2n−1A−(n−1)lA^{n}=2^{n-1}A-(n-1)lAn=2n−1A−(n−1)l

An=nA+(n−1)lA^{n}=nA+(n-1)lAn=nA+(n−1)l

An=2n−1A+(n−1)lA^{n}=2^{n-1}A+(n-1)lAn=2n−1A+(n−1)l

Solution

Question 5

R

N

Q

Z

Solution

Question 6

For all nnn

For all n>1n>1n>1

For all n>mn>mn>m , mmm being a fixed positive integer

Nothing can be said

Solution

Question 7

2n22^{n^2}2n2

2n+42^{n+4}2n+4

2n+22^{n+2}2n+2

2n+32^{n+3}2n+3

Solution

Question 8

131313

111111

111

555

Solution

Question 9

S(1)\mathrm{S}(1)S(1) is correct

S(K)⇒S(K+1)\mathrm{S}(\mathrm{K})\Rightarrow \mathrm{S}(\mathrm{K}+1)S(K)⇒S(K+1)

S(K)≠S(\mathrm{K})\neqS(K)= S(K+++1)

Principle of mathematical induction can be used to prove the formula

Solution

Question 10

x−yx-yx−y

x+yx+yx+y

xyxyxy

x2+y2x^{2}+y^{2}x2+y2

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

P $(5)$ is false but P $(3)$ is true

Both P $(3)$ and P $(5)$ are false

P $(3)$ is false but P $(5)$ is true

Both P $(3)$ and P $(5)$ are true

$S (k)$ $\Rightarrow$ $S (k + 1)$

$S (k)$ ${\nRightarrow}$ $S (k + 1)$

$S (1)$ is correct

$A^{n}=nA-(n-1)l$

$A^{n}=2^{n-1}A-(n-1)l$

$A^{n}=nA+(n-1)l$

$A^{n}=2^{n-1}A+(n-1)l$

For all $n$

For all $n>1$

For all $n>m$ , $m$ being a fixed positive integer

$2^{n^2}$

$2^{n+4}$

$2^{n+2}$

$2^{n+3}$

$13$

$11$

$1$

$5$

$\mathrm{S}(1)$ is correct

$\mathrm{S}(\mathrm{K})\Rightarrow \mathrm{S}(\mathrm{K}+1)$

$S(\mathrm{K})\neq$ S(K $+$ 1)

$x-y$

$x+y$

$xy$

$x^{2}+y^{2}$