Principle of Mathematical Induction and Theorems

Define well ordering principle in mathematical induction.

Prove the following by induction:

$\frac{1}{n+1}+\frac{1}{n+2}+\dots +\frac{1}{3n+1}>1$ for every positive integer $n$.

Define well ordering principle in mathematical induction.

Prove the following by principle of mathematical induction $\forall$ $n\in N$

$1+\frac{1}{1+2}+\frac{1}{1+2+3}+\dots +\frac{1}{1+2+3+\dots +n}=\frac{2n}{n+1}$

Define well ordering principle in mathematical induction.

Prove the following by principle of mathematical induction $\forall n\in N$ 

$\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\dots ..+\frac{1}{n(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}$

Define well ordering principle in mathematical induction.

Prove the following by the principle of mathematical induction for all $n\in N:$

$1.2+2.2^2+3.2^3+\dots \dots +n.2^n=(n-1)2^{n+1}+2$

Define well ordering principle in mathematical induction.

Prove the following by the principle of mathematical induction for all $n\in N:$

$1.3+2.3^2+3.3^3+\dots ..+n.3^n=\frac{(2n-1)3^{n+1}+3}{4}$

Define well ordering principle in mathematical induction.

Prove that $(1+x{)}^n\ge 1+nx,$ for all natural numbers $n,$ where $x>-1$

Let $P\left(n\right)$ be a statement for each natural number $n$. Assume that $P\left(n+1\right)$ is a true statement whenever $P\left(n\right)$ is a true statement. Suppose $P\left(2018\right)$ is true. Then which one of the following statements is true?

The least positive integral value of $\lambda$ such that ${10}^{50}+\left(3\times {\left(4\right)}^{52}\right)+\lambda$ is divisible by $9$ is

A student was asked to prove a statement by induction. He proved $P\left(3\right)$ is true and $P\left(n\right)$ is true $\Rightarrow P(n+1)$ is true, $n\in \mathrm{N}$. On the basis of this, he can conclude that $P\left(n\right)$ is true for:

${\left(1+x\right)}^n-nx-1$ is divisible by (where $n\in N$)

The statement $P\left(n\right):1\times 1!+2\times 2!+3\times 3!+...$$..+n\times n!=\left(n+1\right)!-1$ is

The largest natural number by which $3^{2n}-1, n\in N$ is divisible.

If $A=\left[\begin{array}{ll}1& 0\\ 1& 1\end{array}\right]$ and $I=\left[\begin{array}{ll}1& 0\\ 0& 1\end{array}\right],$ then which one of the following holds, $\forall n\ge 1$ by the principle of mathematical induction?

If $n\mathit{\in }N$ , then $x^{2n-1}+y^{2n-1}$ is divisible by

If $\mathrm{P}\left(n\right):2^n<n!, n\in \mathrm{N},$ then $\mathrm{P}\left(n\right)$ is true for:

The smallest positive integer for which the statement $3^{n+1}<4^n$ holds is

If $n\in N,$ then $2^n$

For natural number $n$, $2^n\left(n-1\right)!<n^n,$ if

Statement I $\left[{\left(m+1\right)}^7-m^7-1\right]$ is divisible by $7$ for each $m\in N.$

Statement II $\left[m^7-m\right]$ is divisible $7$ for each $m\in N.$

When $P$ is a natural number, then $P^{n+1}+{\left(P+1\right)}^{2n-1}$ is divisible by