Neat identity

Once you play a little with the numbers, it's quite easy to notice that no matter what $n$ is equal to, the result is always $\boxed{1}$ . This would mean that $(\displaystyle \sum_{k=1}^n k)^2 = \displaystyle \sum_{k=1}^n k^3$ .

Now let's consider $n=2$ . If we were to increase the side length of the square by $3$ now, it should add as much area to our square as volume to our solid if the above statement is true. That means that the dark blue coloured area should be equal to the volume of the dark blue coloured cube.

Remember that we can write $\displaystyle \sum_{k=1}^n k$ as $\frac {n(n+1)}{2}$ .

This way we can express the dark coloured area as $A = (\frac {n(n+1)}{2})^2 - (\frac {(n-1)n}{2})^2$ .

$(\frac {n(n+1)}{2})^2 - (\frac {(n-1)n}{2})^2 = \frac {(n^2+n)^2-(n^2-n)^2}{4} = \frac {(n^4+2n^3+n^2)-(n^4-2n^3+n^2)}{4} = \frac {4n^3}{4} = n^3$

I don't need to tell you, that the volume of the dark coloured cube is also $n^3$ , so this identity holds true.

The main condition is $n = 27$

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$Side~of~the~square~(s) = \displaystyle \sum_{k = 1}^{n = 27} k = \dfrac{n(n + 1)}{2} = \dfrac{27 \times 28}{2} = 378$

Now, $Area~of~the~square = s^2 = (378)^2$

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To obtain the volume of resulting solid we have to add all the volumes of the individual cubes.

$Volume~of~the~final~solid = 1^3 + 2^3 + 3^3 + ..... + 27^3 \quad (or) \quad \displaystyle \sum_{k = 1}^{n = 27} k^3$

$\implies \dfrac{n^2(n +1)^2}{4} = \dfrac{27^2 \times 28^2}{4} = (378)^2$

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Now, $\text{Ratio} = \dfrac{\text{Area of the square}}{\text{Volume of the solid}} = \dfrac{(378)^2}{(378)^2}$

$\implies 1 : 1$