Question

Find the sum of the series $1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n(n+1)$.

Answer 1

Assume that
$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n(n+1)=A+Bn+C{n}^2+D{n}^3+E{n}^4+\cdots$,
where $A,B,C,D,E,\cdots$ are quantities independent of $n$, whose values have to be determined.
Change $n$ into $n+1$; then
$1\cdot 2+2\cdot 3+\cdots +n(n+1)+(n+1)(n+2)=A+B(n+1)+C{(n+1)}^2+D{(n+1)}^3+E{(n+1)}^4+\cdots$
By subtraction,
$(n+1)(n+2)=B+C(2n+1)+D(3n^2+3n+1)+E(4n^3+6n^2+4n+1)+\cdots$.
This equation being true for all integral values of $n$, the coefficients of the respective powers of $n$ on each side must be equal; thus $E$ and all succeeding coefficients must be equal to zero, and
$3D=1$; $3D+2C=3$; $D+C+B=2$;
whence $D=\frac{1}{3},C=1,B=\frac{2}{3}$.
Hence the sum $=A+\frac{2n}{3}+n^2+\frac{1}{3}{n}^3$.
To find $A$, put $n=1$; the series then reduces to its first term, and
$2=A+2$, or $A=0$.
Hence $1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n(n+1)=\frac{1}{3}n(n+1)(n+2)$.