Correct Answer - Option 1 : 240

__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⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + 4 ⋅ 5 +.....+ 8 ⋅ 9

As we know that, the nth term of the given series is given by T_{n} = n (n + 1)

Sum of series = \(\rm S_n = \sum T_n= \sum (n^2+n)=\sum n^2+\sum n\)

As we know that, \(\rm 1^2+2^2+3^2+4^2+....+ n^2=\sum n^2 = \frac{n(n+1)(2n+1)}{6}\) and \(\rm 1+2+3+4+....+n =\sum n = \frac{n(n+1)}{2}\)

\(⇒ \rm S_n = \frac{n(n+1)(n+2)}{3}\)

Here, n = 8

\(⇒ \rm S_8 = \frac{8 \;\times \;9 \;\times \;10}{3} = 240\)

Hence, **option 1** is the correct answer.