Principles of Mathematical Induction

$1^2+2^2+3^2+...+n^2=?$,  $\forall n\in \mathrm{N}$.

"Every nonempty subset of natural numbers has a minimal element" is an example for Well-ordering Principle

$\frac{3}{4}+\frac{15}{16}+\frac{63}{64}+\dots$ upto $n$  terms is equal to

If $x^n-1$ is divisible by $x-k,$ then the least positive integral value of $k$ is

Consider the following two statements :

I. If $n$ is a composite number, then $n$ divides $\left(n-1\right)!$

II. There are infinitely many natural numbers $n$ such that $n^3+2n^2+n$ divides $n!$

Then

Let $P\left(n\right)$ be a statement and let $P\left(n\right)\to P\left(n+1\right)$ is true for all natural numbers $n$, then $P\left(n\right)$ is true

If $2\left(4^{2n+1}\right)+3^{3n+1}$ is divisible by $k, k>1$ for all $n\in \mathrm{\mathbb{N}}$ then the value of $k$ is

$x^n+y^n$ is divisible by

If $1.2+2.3+3.4+\dots .+n(n+1)=\frac{n(n+1)(n+2)}{3}=P\left(n\right)$, then 

If  $3.6+6.9+9.12+\dots ..+3n(3n+3)$$=3n(n+1)(n+2)=P\left(n\right)$ then 

For all $n\in N,$ $n(n+1)(n+5)$ is a multiple of 

Inequality $3^n<(n+1)!, n\in N$

For all $n\in N$,  $2.5+5.8+8.11+\dots .+(3n-1)(3n+2)=n\left(3n^2+6n+1\right)=P\left(n\right)$ then 

If we take any three consecutive natural numbers, then the sum of their cubes is always divisible by $$

For every natural number $n$,  $n^3+(n+1{)}^3+(n+2{)}^3$ is divisible by  

For all $n\in N,$ $1·2+2·2^2+3·2^3+\dots \dots +n·2^n=$

For all $n\in N,$ $1.3+2.3^2+3.3^3+\dots ..+n.3^n=$

The statement $:2^n>3n$, $n\in N$.

For all $n\in N$, $7^{2n}+2^{3n-3}\cdot 3^{n-1}$ is divisible by $$ 

$\forall n\in N, 6^{n+2}+7^{2n+1}$ is divisible by