Concept:

• Sum of the first n Natural Numbers \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)
• Sum of the Square of the first n Natural Numbers \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\)
• Sum of the Cubes of the first n Natural Numbers \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

Calculation:

Here, we have to find the sum of the series (1 × 0 × 2) + (2 × 1 × 3) + (3 × 2 × 4) + (4 × 3 × 5) +....up to n terms.

nth term of the series = T_n = n ⋅ (n - 1) ⋅ (n + 1) = n3 - n 

Sum of series = \(\rm S_n = \sum T_n= \sum (n^3-n)=\sum n^3-\sum n\)

As we know that, \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)  and \(\rm 1^3+2^3+3^3+4^3+....+ n^3=\sum n^3 = \frac{[n(n+1)]^2}{4}\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )^2-\frac{n(n+1)}{2}\)

\(\rm \Rightarrow S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{n(n+1)}{2}-1 \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)}{2}} \right )\left ( \frac{(n-1)(n+2)}{2} \right )\)

\(\Rightarrow \rm S_n =\left ( { \frac{n(n+1)(n-1)(n+2)}{4}} \right )\)

\(\Rightarrow \rm S_n = { \frac{n(n^2-1)(n+2)}{4}} \)

Hence, option 1 is the correct answer.