How can you prove that 1+5+9+⋯+(4n−3)=2n2−n1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} – n without using induction?

Using mathematical induction, I have proved that

1+5+9++(4n3)=2n2n

for every integer n>0.

I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?

Here is the way I have solved this using PMI.

Base Case: since 1=2·121, the formula holds for n=1.

Assuming that the
formula holds for some integer k1, that is,

1+5+9++(4k3)=2k2k

I show that

1+5+9++[4(k+1)3]=2(k+1)2(k+1).

Now if I use hypothesis I observe.

1+5+9++[4(k+1)3]=[1+5+9++(4k3)]+4(k+1)3=(2k2k)+(4k+1)=2k2+3k+1=2(k+1)2(k+1)

Answer

Follow the rainbow!

\Large \color{PaleVioletRed}1 + \color{DarkViolet}5 + \color{DodgerBlue}9 + \dots + \color{LightCoral}{(4n-3)} = n(2n-1) = 2n^2-n

enter image description here

Attribution
Source : Link , Question Author : Chan C , Answer Author : Lynn

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