Question

Sum the series
$1.2+2.3+3.4+...$ to $n$ terms.

Answer 1

The $n^{th}$ term $=n(n+1)=n^2+n$; and by writing down each term in a similar form we shall have two columns, one consisting of the first $n$ natural numbers, and the other of their squares.
$\therefore$ the sum $=\sum n^2+\sum n$
$=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$
$=\frac{n(n+1)}{2}\{\frac{2n+1}{3}+1\}$
$=\frac{n(n+1)(n+2)}{3}$.