1+5+9+...+(4m-3)=n(2n-1)

Question

for my friend

Answer 1

1+5+9 +... is a sequens av positiv integers with start=1 & step size=4

last number=1+stepsize*(n-1)

averaej=(start number +last number)/n

sum=n*averaej/2

for stepsize=4, last=1+4*(n-1)

averaej=2+4*(n-1)

sum=n*[1+2*(n-1)]  or n*(2n-1)

zampel...n=5: sum=5*9=45