Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 124106 by mathocean1 last updated on 30/Nov/20

U_(n ) is a sequence of real numbers   defined by U_0 =0 and for n ∈ N,   U_(n+1) =(√(U_n +6))  1. show that 0≤U_n ≤3.  2. show that U_(n ) is non−decreasing.  3. show that 3−U_(n+1) ≤((3−U_n )/3)  4.  Deduct that 0≤3−U_(n+1) ≤((1/3))^n

$${U}_{{n}\:} {is}\:{a}\:{sequence}\:{of}\:{real}\:{numbers}\: \\ $$$${defined}\:{by}\:{U}_{\mathrm{0}} =\mathrm{0}\:{and}\:{for}\:{n}\:\in\:\mathbb{N},\: \\ $$$${U}_{{n}+\mathrm{1}} =\sqrt{{U}_{{n}} +\mathrm{6}} \\ $$$$\mathrm{1}.\:\mathrm{show}\:\mathrm{that}\:\mathrm{0}\leqslant{U}_{{n}} \leqslant\mathrm{3}. \\ $$$$\mathrm{2}.\:{show}\:{that}\:{U}_{{n}\:} {is}\:{non}−{decreasing}. \\ $$$$\mathrm{3}.\:{show}\:{that}\:\mathrm{3}−{U}_{{n}+\mathrm{1}} \leqslant\frac{\mathrm{3}−{U}_{{n}} }{\mathrm{3}} \\ $$$$\mathrm{4}.\:\:{Deduct}\:{that}\:\mathrm{0}\leqslant\mathrm{3}−{U}_{{n}+\mathrm{1}} \leqslant\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com