Question Number 25283 by Tinkutara last updated on 07/Dec/17
Commented by Tinkutara last updated on 07/Dec/17
$${Solve}\:{for}\:{x}. \\ $$
Answered by naka3546 last updated on 07/Dec/17
$${x}\:\:=\:\:\mathrm{72} \\ $$
Answered by ajfour last updated on 08/Dec/17
$${Let}\:{S}=\frac{\mathrm{3}}{\mathrm{1}.\mathrm{2}.\mathrm{4}}+\frac{\mathrm{4}}{\mathrm{2}.\mathrm{3}.\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{3}.\mathrm{4}.\mathrm{6}}+…. \\ $$$${S}=\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}{T}_{{r}} \\ $$$${T}_{{r}} =\frac{{r}+\mathrm{2}}{{r}\left({r}+\mathrm{1}\right)\left({r}+\mathrm{3}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}{r}}\left(\frac{\mathrm{1}}{{r}+\mathrm{1}}+\frac{\mathrm{1}}{{r}+\mathrm{3}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{r}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{{r}}−\frac{\mathrm{1}}{{r}+\mathrm{3}}\right) \\ $$$${S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{r}}−\underset{{r}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{r}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{6}}\left(\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{r}}−\underset{{r}=\mathrm{4}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{r}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right)\: \\ $$$${S}=\frac{\mathrm{58}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{11}}{\mathrm{36}}=\frac{\mathrm{29}}{\mathrm{36}} \\ $$$$\Rightarrow\:\:{x}=\mathrm{72}\:. \\ $$
Commented by Tinkutara last updated on 08/Dec/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$