Prove by the Principle of Mathematical Induction: $1+5+9+...+(4n-3)=n(2n-1)$ for all natural numbers n.

Assume given statement

Let given statement is P(n),

i.e., $P\left(n\right):1+5+9+...+(4n-3)=n(2n-1)\forall nϵN$

Check that statement is true for $n=1$

The given statement is true for $n=1$

As $LHS=1$

and$RHS=1\left(2\right(1)-1)=1(2-1)=1=LHS$

Assume P(k) to be true and then prove $P(k+1)$ is true.

Assume that P(n) is true for$n=k$

$\Rightarrow 1+5+9+....(4k-3)=k(2k-1)\dots \left(1\right)$

To prove: $1+5+9+....+\left(4\right(k+1)-3)$

$=(k+1)\left(2\right(k+1)-1)$

$LHS=1+5+9+...+(4k-3)+\left({}_{-3}^{4(k+1)}\right)$

$=k(2k-1)+4(k+1)-3$

$=2k^2-k+4k+4-3$

$=2k^2+3k+1$

$=2k^2+2k+k+1$

$=2k(k+1)+1(k+1)$

$=(k+1)(2k+1)$

$=(k+1)[2k+2-1]$

$=(k+1)\left[2\right(k+1)-1]=RHS$

Thus, $P(k+1)$ is true whenever P(k) is true.

Hence, By Principle of mathematical Induction P(n) is true for all natural numbers n.