Correct Answer: 3080

Solution : Given: The series is $1 \times 2+2 \times 3+3 \times 4+4 \times 5+\ldots \ldots$.
The sum of the first $n$ consecutive number = $\frac{n(n+1)}{2}$
The sum of the square of the first $n$ consecutive number = $\frac{n(n + 1) (2n + 1)}{6}$
The sum of the first 20 terms of the following series is given as,
$n$ the term = $n(n+1)$
Sum = $\sum n(n+1)=\sum (n^2+n)=\sum n^2+\sum n$
Sum = $\frac{n(n + 1) (2n + 1)}{6}+\frac{n(n+1)}{2}$
⇒ Sum = $\frac{n(n+1)}{2}[\frac{2n+1}{3}+1]$
⇒ Sum = $\frac{n(n+1)}{2}[\frac{2n+1+3}{3}]$
⇒ Sum = $\frac{n(n+1)(2n+4)}{6}$ (equation 1)
Substitute $n=20$ in the equation (1),
Sum = $\frac{20(20+1)(2\times20+4)}{6}$
⇒ Sum = $\frac{20\times 21\times44}{6}$ = $3080$
Hence, the correct answer is 3080.