Principle of Mathematical Induction Test - 9

$$ \displaystyle P\left( n \right) =\frac { 1 }{ \sqrt { 1 }  } +\frac { 1 }{ \sqrt { 2 }  } +...+\frac { 1 }{ \sqrt { n }  } $$ 
$$ \displaystyle P\left( 2 \right) =\frac { 1 }{ \sqrt { 1 }  } +\frac { 1 }{ \sqrt { 2 }  } >\sqrt { 2 } $$
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. 

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

Putting $$k=1$$, we get L.H.S=1 and R.H.S=4. Hence $$A$$ and $$D$$ are incorrect.

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

Mathematical induction is a method of mathematical proof typically used to establish a given statement for all natural numbers.

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 $$\forall k$$ $$\in$$ $$N$$ 

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 $$m$$.

$$5^n-2^n$$ 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

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

$$\displaystyle P\left ( n \right )=x^{2n-1}+y^{2n-1}\forall n \epsilon N.
$$ 
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