Question

Find the sum to n terms of the series $\frac{3}{\mathrm{1.2.4}}+\frac{4}{\mathrm{2.3.5}}+\frac{5}{\mathrm{3.4.6}}+....$$

Answer 1

Here the rule is not directly applicable, because although $1,2,3,....,$ the first factors of the several denominators, are in arithmetical progression, the factors of any one denominator are not. In this example we may proceed as follows:
$u_n=\frac{n+2}{n(n+1)(n+3)}=\frac{(n+2{)}^2}{n(n+1)(n+2)(n+3)}$
$=\frac{n(n+1)+3n+4}{n(n+1)(n+2)(n+3)}$
$=\frac{1}{(n+2)(n+3)}+\frac{3}{(n+1)(n+2)(n+3)}+\frac{4}{n(n+1)(n+2)(n+3)}.$
Each of these expressions may now be taken as the $n^{th}$ term of a series to which the rule is applicable.
$\therefore S_n=C-\frac{1}{n+3}-\frac{3}{2(n+2)(n+3)}-\frac{4}{3(n+1)(n+2)(n+3)};$
Put $n=1,$ then
$\frac{3}{\mathrm{1.2.4}}=C-\frac{1}{4}-\frac{3}{\mathrm{2.3.4}}-\frac{4}{\mathrm{3.2.3.4}};$ hence $C=\frac{29}{36};$
$\therefore S_n=\frac{29}{36}-\frac{1}{n+3}-\frac{3}{2(n+2)(n+3)}-\frac{4}{3(n+1)(n+2)(n+3)}$.