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Question Number 7743 by Tawakalitu. last updated on 13/Sep/16

All the terms of the arithmetic progession u_1 , u_2 , u_3 , ... u_n are positive . use induction to prove that for n ≥ 2 (1/(u_1 u_2 )) + (1/(u_2 u_3 )) + (1/(u_3 u_4 )) + ... (1/(u_(n − 1) u_n )) = ((n − 1)/(u_1 u_n ))

$${All}\:{the}\:{terms}\:{of}\:{the}\:{arithmetic}\:{progession}\: \\ $$$${u}_{\mathrm{1}} ,\:{u}_{\mathrm{2}} ,\:{u}_{\mathrm{3}} ,\:...\:{u}_{{n}} \:\:{are}\:{positive}\:.\:{use}\:{induction}\:{to} \\ $$$${prove}\:{that}\:{for}\:{n}\:\geqslant\:\mathrm{2} \\ $$$$\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{{u}_{\mathrm{2}} {u}_{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{{u}_{\mathrm{3}} {u}_{\mathrm{4}} }\:+\:...\:\frac{\mathrm{1}}{{u}_{{n}\:−\:\mathrm{1}} {u}_{{n}} }\:\:=\:\:\frac{{n}\:−\:\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{n}} } \\ $$

Answered by Yozzia last updated on 13/Sep/16

Let P(n) denote the statement, ∀n∈N,n≥2, (1/(u_1 u_2 ))+(1/(u_2 u_3 ))+(1/(u_3 u_4 ))+...+(1/(u_(n−1) u_n ))=((n−1)/(u_1 u_n )) for u_1 ,u_2 ,...,u_n being an A.P with all terms positive. For n=2, P(2) says (1/(u_1 u_2 ))=((2−1)/(u_1 u_2 ))=(1/(u_1 u_2 )) which is true. Assume P(n) is true for some n=k,i.e Σ_(i=1) ^(k−1) (1/(u_i u_(i+1) ))=((k−1)/(u_1 u_k )). Σ_(i=1) ^(k−1) (1/(u_i u_(i+1) ))+(1/(u_k u_(k+1) ))=((k−1)/(u_1 u_k ))+(1/(u_k u_(k+1) )) Σ_(i=1) ^k (1/(u_i u_(i+1) ))=(1/u_k )(((k−1)/u_1 )+(1/u_(k+1) )) Σ_(i=1) ^k (1/(u_i u_(i+1) ))=(1/u_k )((((k−1)u_(k+1) +u_1 )/(u_1 u_(k+1) ))) For the A.P, u_t =u_1 +(t−1)d for t∈N, d=common difference. ∴ u_(k+1) =u_1 +kd Σ_(i=1) ^k (1/(u_i u_(i+1) ))=(1/u_k )((((k−1)(u_1 +kd)+u_1 )/(u_1 u_(k+1) ))) Σ_(i=1) ^k (1/(u_i u_(i+1) ))=(1/u_k )(((k(u_1 +kd)−u_1 −kd+u_1 )/(u_1 u_(k+1) ))) Σ_(i=1) ^k (1/(u_i u_(i+1) ))=(1/u_k )(((k(u_1 +kd−d))/(u_1 u_(k+1) )))=(1/u_k )×((k(u_1 +(k−1)d))/(u_1 u_(k+1) )) Σ_(i=1) ^k (1/(u_i u_(i+1) ))=(1/u_k )×((ku_k )/(u_1 u_(k+1) ))=(k/(u_1 u_(k+1) )) which is P(k+1) Therefore, if P(k) is true, then P(k+1) is true. Since P(2) is true then, by P.M.I, P(n) is true ∀n∈N,n≥2.

$${Let}\:{P}\left({n}\right)\:{denote}\:{the}\:{statement},\:\forall{n}\in\mathbb{N},{n}\geqslant\mathrm{2}, \\ $$$$\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }+\frac{\mathrm{1}}{{u}_{\mathrm{2}} {u}_{\mathrm{3}} }+\frac{\mathrm{1}}{{u}_{\mathrm{3}} {u}_{\mathrm{4}} }+...+\frac{\mathrm{1}}{{u}_{{n}−\mathrm{1}} {u}_{{n}} }=\frac{{n}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{n}} } \\ $$$${for}\:{u}_{\mathrm{1}} ,{u}_{\mathrm{2}} ,...,{u}_{{n}} \:{being}\:{an}\:{A}.{P}\:{with}\:{all}\:{terms}\:{positive}. \\ $$$${For}\:{n}=\mathrm{2},\:{P}\left(\mathrm{2}\right)\:{says}\:\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }=\frac{\mathrm{2}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} }=\frac{\mathrm{1}}{{u}_{\mathrm{1}} {u}_{\mathrm{2}} } \\ $$$${which}\:{is}\:{true}. \\ $$$${Assume}\:{P}\left({n}\right)\:{is}\:{true}\:{for}\:{some}\:{n}={k},{i}.{e} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{{k}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{k}} }. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }+\frac{\mathrm{1}}{{u}_{{k}} {u}_{{k}+\mathrm{1}} }=\frac{{k}−\mathrm{1}}{{u}_{\mathrm{1}} {u}_{{k}} }+\frac{\mathrm{1}}{{u}_{{k}} {u}_{{k}+\mathrm{1}} } \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{{k}−\mathrm{1}}{{u}_{\mathrm{1}} }+\frac{\mathrm{1}}{{u}_{{k}+\mathrm{1}} }\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{\left({k}−\mathrm{1}\right){u}_{{k}+\mathrm{1}} +{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right) \\ $$$${For}\:{the}\:{A}.{P},\:{u}_{{t}} ={u}_{\mathrm{1}} +\left({t}−\mathrm{1}\right){d}\:{for}\:{t}\in\mathbb{N},\:{d}={common}\:{difference}. \\ $$$$\therefore\:{u}_{{k}+\mathrm{1}} ={u}_{\mathrm{1}} +{kd} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{\left({k}−\mathrm{1}\right)\left({u}_{\mathrm{1}} +{kd}\right)+{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{{k}\left({u}_{\mathrm{1}} +{kd}\right)−{u}_{\mathrm{1}} −{kd}+{u}_{\mathrm{1}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right) \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }\left(\frac{{k}\left({u}_{\mathrm{1}} +{kd}−{d}\right)}{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\right)=\frac{\mathrm{1}}{{u}_{{k}} }×\frac{{k}\left({u}_{\mathrm{1}} +\left({k}−\mathrm{1}\right){d}\right)}{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} } \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{u}_{{i}} {u}_{{i}+\mathrm{1}} }=\frac{\mathrm{1}}{{u}_{{k}} }×\frac{{ku}_{{k}} }{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }=\frac{{k}}{{u}_{\mathrm{1}} {u}_{{k}+\mathrm{1}} }\:{which}\:{is}\:{P}\left({k}+\mathrm{1}\right) \\ $$$${Therefore},\:{if}\:{P}\left({k}\right)\:{is}\:{true},\:{then}\:{P}\left({k}+\mathrm{1}\right)\:{is}\:{true}. \\ $$$${Since}\:{P}\left(\mathrm{2}\right)\:{is}\:{true}\:{then},\:{by}\:{P}.{M}.{I},\: \\ $$$${P}\left({n}\right)\:{is}\:{true}\:\forall{n}\in\mathbb{N},{n}\geqslant\mathrm{2}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 13/Sep/16

Wow, Thank you so much.

$${Wow},\:{Thank}\:{you}\:{so}\:{much}. \\ $$

Commented by Tawakalitu. last updated on 13/Sep/16

I really appreciate

$${I}\:\:{really}\:{appreciate} \\ $$

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