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Question-188657




Question Number 188657 by Ajetunmobi last updated on 04/Mar/23
Answered by Ajetunmobi last updated on 04/Mar/23
please solve with detail solution
$${please}\:{solve}\:{with}\:{detail}\:{solution} \\ $$
Answered by witcher3 last updated on 05/Mar/23
U_(n+1) −U_n =(√(u_n +2))−(√(u_(n−1) +2))  =((u_n −u_(n−1) )/( (√(u_n +2))+(√(u_(n−1) +2))))...
$$\mathrm{U}_{\mathrm{n}+\mathrm{1}} −\mathrm{U}_{\mathrm{n}} =\sqrt{\mathrm{u}_{\mathrm{n}} +\mathrm{2}}−\sqrt{\mathrm{u}_{\mathrm{n}−\mathrm{1}} +\mathrm{2}} \\ $$$$=\frac{\mathrm{u}_{\mathrm{n}} −\mathrm{u}_{\mathrm{n}−\mathrm{1}} }{\:\sqrt{\mathrm{u}_{\mathrm{n}} +\mathrm{2}}+\sqrt{\mathrm{u}_{\mathrm{n}−\mathrm{1}} +\mathrm{2}}}… \\ $$
Answered by mehdee42 last updated on 05/Mar/23
∀n∈N , u_n =2 ⇒ ∀i∈N ; u_i −u_(i−1) =0   !!!?
$$\forall{n}\in\mathbb{N}\:,\:{u}_{{n}} =\mathrm{2}\:\Rightarrow\:\forall{i}\in\mathbb{N}\:;\:{u}_{{i}} −{u}_{{i}−\mathrm{1}} =\mathrm{0}\: \\ $$$$!!!? \\ $$

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