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Question Number 113059 by bemath last updated on 11/Sep/20

Answered by john santu last updated on 11/Sep/20

$$\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}=$$$$\underset{k=1}{\sum ^4}\frac{1}{(k+1).(k+2).(k+3)}$$$$\frac{1}{(k+1)(k+2)(k+3)}=\frac{p}{k+1}+\frac{q}{k+2}+\frac{r}{k+3}$$$$1=p(k+2)(k+3)+q(k+1)(k+3)+r(k+1)(k+2)$$$$k=-1\Rightarrow 1=p\left(1\right)\left(2\right);p=\frac{1}{2}$$$$k=-2\Rightarrow 1=q(-1)\left(1\right);q=-1$$$$k=-3\Rightarrow 1=r(-2)(-1);r=\frac{1}{2}$$$$sowehave\underset{k=1}{\sum ^4}\frac{1}{2(k+1)}+\frac{1}{2(k+3)}-\frac{1}{(k+2)}$$$$=\frac{1}{4}+\frac{1}{8}-\frac{1}{3}+\frac{1}{6}+\frac{1}{10}-\frac{1}{4}+\frac{1}{8}+\frac{1}{12}-\frac{1}{5}$$$$+\frac{1}{10}+\frac{1}{14}-\frac{1}{6}$$$$=\frac{1}{4}+\frac{1}{12}+\frac{1}{14}-\frac{1}{3}=\frac{7+6}{84}-\frac{1}{12}=\frac{13}{84}-\frac{7}{84}$$$$=\frac{6}{84}=\frac{1}{14}$$

Answered by floor(10²Eta[1]) last updated on 11/Sep/20

$$\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}$$$$=\frac{1}{4!}+\frac{2!}{5!}+\frac{3!}{6!}+\frac{4!}{7!}$$$$=\frac{1}{4!}(1+\frac{2!}{5}+\frac{3!}{6.5}+\frac{4!}{7.6.5})$$$$=\frac{1}{4!}(1+\frac{2}{5}+\frac{1}{5}+\frac{4}{7.5})$$$$=\frac{1}{4!}\left(\frac{60}{5.7}\right)=\frac{1}{4}\times \frac{10}{5.7}=\frac{1}{2}\times \frac{1}{7}=\frac{1}{14}$$

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