It's About RHS and LHS

Note that the number of terms in each row is in an Arithmetic Progression with $a=3$ and $d=2$ .
The last term of RHS of the $n^{th}$ row is the sum of the first $n$ terms of this AP.

We want the last term of RHS of the $80^{th}$ row; which is; by using the formula for the sum of an AP;
$\frac{80}{2}((2)(3) + (79)(2)) =40*164 = \boxed {6560}$

Lemma : $n^2$ is equal to the sum of the first $|n|$ positive odd integers.

Proof : $x^2 - y^2 = (x+y)(x-y)$ .

WLOG $x > y$ .

If $x-y=1$ , then $x^2 - y^2 = 2y+1$ .

Now we prove the statement using induction:

$1^2 = 1$

$(n+1)^2 = n^2 + 2n + 1$ .

Solution : As each row contains $2$ more integers than the last, and the sequence begins with $3$ integers in a row, the last number on the $RHS$ of the $n$ th row is equal to $(n+1)^2 -1)$ .

So the answer is $81^2 - 1 = 6560$ .

Its easy to note that the first number on the $LHS$ of any row $n$ is given by $n^2$

since the terms are consecutive the last term on the $RHS$ of the $n$ th row should be $(n+1)^2-1$

applying this we get last term on the $80$ th row = $81^2-1=\boxed{6560}$

The last number in every row on the $LHS$ makes a sequence $2, 6, 12, 20, ....$

The last number of the $80^{th}$ row on the $LHS$ is

$80(80+1) = 80 \times 81$

$= 6480$ .

On the $80^{th}$ row, $LHS$ has $81$ terms, and $RHS$ has $80$ terms

The first number of the $80^{th}$ on $RHS$ is $6481$ .

Then, the last number of the $80^{th}$ on $RHS$ is $6481+79 = 6560$ .