$\sum_{j=1}^n(3j - 2) = \frac{n(3n - 1)}{2} $ using geometry

Question

I know how to prove it by induction, but is there a way using geometry? I tried to use triangles to get the half but can't proceed.

For other proofs than induction see this post. – Dietrich Burde May 09 '20 at 18:05 — Dietrich Burde, May 09 '20 at 18:05

score 4 · Accepted Answer · answered May 09 '20 at 18:06

Imagine that you have $n$ rectangles of dimensions $1\times(3j-2)$ for $j=1,2,\ldots,n$. Arrange them in a sort of staircase, like this for $n=3$:

*
****
*******

Now make a second copy, but upside down:

*******
   ****
      *

Finally, slide the two together to make an $n\times(3n-1)$ rectangle:

*|*******
****|****
*******|*

Here I’ve left vertical bars between the two ‘staircases’ to make the construction clear, but of course they should be removed.

score 1 · Answer 2 · answered May 09 '20 at 18:17

Not completely geometric.. but without Induction

$$ \sum_{j=1}^n(3j - 2) = \sum_{j=1}^n 3j - \sum_{j=1}^n 2 = 3\sum_{j=1}^n j - 2\sum_{j=1}^n 1 $$ Now, $\sum_{j=1}^n 1$ is $n$ and $$ \sum_{j=1}^n j $$ could be represented in geometry, for example for n=3 i.e. 1 + 2 + 3

***
**
*

which is half of square of $3X3$ (*s) + half of # of diagonal elements (as they are considered half while considering identity as half of square).

Hence,
$$ \sum_{j=1}^n j = \frac{n^2}{2} + \frac{n}{2} =\frac{n(n+1)}{2} $$

Finally, $$ \sum_{j=1}^n(3j - 2) = \frac{3n(n+1)}{2} - 2n = \frac{n(3n - 1)}{2} $$

Thanks, I like this answer too, I'll give you a +1. But I will accept the other one because is more of what I was looking for. — Mystery, May 09 '20 at 18:52

$\sum_{j=1}^n(3j - 2) = \frac{n(3n - 1)}{2} $ using geometry

2 Answers2