Correct Answer - Option 2 : 2 only

**Concept:**

**Sum of cubes of natural numbers:**

The sum of the cubes of first n natural numbers is given by:

\(\rm \displaystyle\sum_{k=1}^{n}k^3 = \left[\dfrac{n(n+1)}{2}\right]^2\)

**Sum of squares of natural numbers:**

The sum of squares of first n natural numbers is given by:

\(\rm \displaystyle\sum_{k=1}^{n}k^2 = \left[\dfrac{n(n+1)(2n+1)}{6}\right]\)

**Calculation:**

The sum of cubes of the first 20 natural numbers is given by:

\(\begin{align*} \left[\dfrac{n(n+1)}{2}\right]^2 &= \left[\dfrac{20(20+1)}{2}\right]^2\\ &= (210)^2\\ &=44100 \end{align*}\)

Similarly, the sum of squares of the first 20 natural numbers is given by:

\(\begin{align*} \left[\dfrac{n(n+1)(2n+1)}{6}\right] &= \dfrac{20(20+1)(40+1)}{6}\\ &= \dfrac{20\times 21\times41}{6}\\ &= 2870 \end{align*}\)

Therefore, sum of cubes of the first 20 natural numbers is 44100 and that of squares of the first 20 natural numbers is 2870.

Thus only the second statement is correct.