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Question

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


Solution

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:
$$u_n=\displaystyle\frac{n+2}{n(n+1)(n+3)}=\frac{(n+2)^2}{n(n+1)(n+2)(n+3)}$$
$$=\displaystyle\frac{n(n+1)+3n+4}{n(n+1)(n+2)(n+3)}$$
$$=\displaystyle\frac{1}{(n+2)(n+3)}+\frac{3}{(n+1)(n+2)(n+3)}+\frac{4}{n(n+1)(n+2)(n+3)}.$$
Each of these expressions may now be taken as the $$n^{th}$$ term of a series to which the rule is applicable.
$$\therefore S_n=C-\displaystyle\frac{1}{n+3}-\frac{3}{2(n+2)(n+3)}-\frac{4}{3(n+1)(n+2)(n+3)};$$
Put $$n=1,$$ then
$$\displaystyle\frac{3}{1.2.4}=C-\frac{1}{4}-\frac{3}{2.3.4}-\frac{4}{3.2.3.4};$$ hence $$C=\displaystyle\frac{29}{36};$$
$$\therefore S_n=\displaystyle\frac{29}{36}-\frac{1}{n+3}-\frac{3}{2(n+2)(n+3)}-\frac{4}{3(n+1)(n+2)(n+3)}$$.

Mathematics

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