Prove the following by induction on n
1+5+9+....+(4n-3) = n(2n-1) for all n => 1
check base case n=1
1=1(2(1)-1) true
assume n=k
$\displaystyle 1+5+9+....+(4k-3)=k(2k-1)$
Show k+1
$\displaystyle 1+5+9+....+(4k-3)+[4(k+1)-3]=$
by the induction hypothsis
$\displaystyle k(2k-1)+[4(k+1)-3]=2k^2-k+4k+4-3=2k^2+3k+1=$
$\displaystyle (k+1)(2k+1)=(k+1)(2(k+1)-1)$
The last one is k+1
QED