Sum of cubes of first 2 positive integers
$$1^3+2^3=1+8=9=3^2$$

Sum of the for the first three positive integers.
$$1^3+2^3+3^3=36=6^3$$

The sum of the first four positive integers.
$$1^3+2^3+3^3+4^3=100=10^2$$

The sum of the nth term of =
$$\left[\frac{2n}{4}\left(n+1\right)\right]^2$$

Test for the second term where n=2
$$\left[\frac{\left(2\cdot2\right)}{4}\left(2+1\right)\right]=3^2$$

Test for third term when n=3
$$\left[\frac{\left(2\cdot3\right)}{4}\cdot\left(3+1\right)\right]^2=\left(\frac{24}{4}\right)^2=6^2$$

Test for the fourth term when n=4
$$\left[\frac{\left(2\cdot4\right)}{4}\cdot\left(4+1\right)\right]^2=\left(\frac{40}{4}\right)^2=10^2$$

For the 10th term when n =10
$$\left[\frac{\left(2\cdot10\right)}{4}\cdot\left(10+1\right)\right]^2=\left(\frac{220}{4}\right)^2=55^2$$

$$answer\ is\ Option\ C$$