Question Number 7743 by Tawakalitu. last updated on 13/Sep/16
$${All}\:{the}\:{terms}\:{of}\:{the}\:{arithmetic}\:{progession}\: \\ $$$${u}_{\mathrm{1}} ,\:{u}_{\mathrm{2}} ,\:{u}_{\mathrm{3}} ,\:…\:{u}_{{n}} \:\:{are}\:{positive}\:.\:{use}\:{induction}\:{to} \\ $$$${prove}\:{that}\:{for}\:{n}\:\geqslant\:\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{u}_{\mathrm{2}} {u}_{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{u}_{\mathrm{3}} {u}_{\mathrm{4}} }\:+\:…\:\frac{\mathrm{1}}{{u}_{{n}\:−\:\mathrm{1}} {u}_{{n}} }\:\:=\:\:\frac{{n}\:−\:\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{n}} } \\ $$
Answered by Yozzia last updated on 13/Sep/16
$${Let}\:{P}\left({n}\right)\:{denote}\:{the}\:{statement},\:\forall{n}\in\mathbb{N},{n}\geqslant\mathrm{2}, \\ $$$$\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }+\frac{\mathrm{1}}{{u}_{\mathrm{2}} {u}_{\mathrm{3}} }+\frac{\mathrm{1}}{{u}_{\mathrm{3}} {u}_{\mathrm{4}} }+…+\frac{\mathrm{1}}{{u}_{{n}−\mathrm{1}} {u}_{{n}} }=\frac{{n}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{n}} } \\ $$$${for}\:{u}_{\mathrm{1}} ,{u}_{\mathrm{2}} ,…,{u}_{{n}} \:{being}\:{an}\:{A}.{P}\:{with}\:{all}\:{terms}\:{positive}. \\ $$$${For}\:{n}=\mathrm{2},\:{P}\left(\mathrm{2}\right)\:{says}\:\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }=\frac{\mathrm{2}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }=\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} } \\ $$$${which}\:{is}\:{true}. \\ $$$${Assume}\:{P}\left({n}\right)\:{is}\:{true}\:{for}\:{some}\:{n}={k},{i}.{e} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{{k}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{k}} }. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }+\frac{\mathrm{1}}{{u}_{{k}} {u}_{{k}+\mathrm{1}} }=\frac{{k}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{k}} }+\frac{\mathrm{1}}{{u}_{{k}} {u}_{{k}+\mathrm{1}} } \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{{k}−\mathrm{1}}{{u}_{\mathrm{1}} }+\frac{\mathrm{1}}{{u}_{{k}+\mathrm{1}} }\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{\left({k}−\mathrm{1}\right){u}_{{k}+\mathrm{1}} +{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right) \\ $$$${For}\:{the}\:{A}.{P},\:{u}_{{t}} ={u}_{\mathrm{1}} +\left({t}−\mathrm{1}\right){d}\:{for}\:{t}\in\mathbb{N},\:{d}={common}\:{difference}. \\ $$$$\therefore\:{u}_{{k}+\mathrm{1}} ={u}_{\mathrm{1}} +{kd} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{\left({k}−\mathrm{1}\right)\left({u}_{\mathrm{1}} +{kd}\right)+{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{{k}\left({u}_{\mathrm{1}} +{kd}\right)−{u}_{\mathrm{1}} −{kd}+{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{{k}\left({u}_{\mathrm{1}} +{kd}−{d}\right)}{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right)=\frac{\mathrm{1}}{{u}_{{k}} }×\frac{{k}\left({u}_{\mathrm{1}} +\left({k}−\mathrm{1}\right){d}\right)}{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} } \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }×\frac{{ku}_{{k}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }=\frac{{k}}{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\:{which}\:{is}\:{P}\left({k}+\mathrm{1}\right) \\ $$$${Therefore},\:{if}\:{P}\left({k}\right)\:{is}\:{true},\:{then}\:{P}\left({k}+\mathrm{1}\right)\:{is}\:{true}. \\ $$$${Since}\:{P}\left(\mathrm{2}\right)\:{is}\:{true}\:{then},\:{by}\:{P}.{M}.{I},\: \\ $$$${P}\left({n}\right)\:{is}\:{true}\:\forall{n}\in\mathbb{N},{n}\geqslant\mathrm{2}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawakalitu. last updated on 13/Sep/16
$${Wow},\:{Thank}\:{you}\:{so}\:{much}. \\ $$
Commented by Tawakalitu. last updated on 13/Sep/16
$${I}\:\:{really}\:{appreciate} \\ $$