Answer
The area for $G_{i}=i^{3}$
The area of ABCD is $\sum_{i=1}^ni^3=[\frac{n(n+1)}{2}]^2$
Work Step by Step
Area for $G_{i}=[\frac{i(i+1)}{2}]^{2}-[\frac{i(i-1)}{2}]^{2}$
$=\frac{i^{2}(i+1)^{2}}{4}-\frac{i^{2}(i-1)^{2}}{4}$
$=\frac{1}{4}i^{2}[(i+1)^{2}-(i-1)^{2}]$
$=\frac{1}{4}i^{2}[4i]$
Hence, the area for $G_{i}=i^{3}$
So the area of ABCD is $\sum_{i=1}^ni^3=[\frac{n(n+1)}{2}]^2$
