Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 72671 by aliesam last updated on 31/Oct/19

Commented by kaivan.ahmadi last updated on 31/Oct/19

a)  ∣x_(n+1) −x_n ∣≤c∣x_n −x_(n−1) ∣≤c^2 ∣x_(n−1) −x_(n−2) ∣≤...≤c^n ∣x_1 −x_0 ∣  b)  ∣x_(n+1) −x_0 ∣=∣(x_(n+1) −x_n )+(x_n −x_(n−1) )+...+(x_1 −x_0 )∣≤  ∣x_(n+1) −x_n ∣+∣x_n −x_(n−1) ∣+...+∣x_1 −x_0 ∣≤  c^n ∣x_1 −x_0 ∣+c^(n−1) ∣x_1 −x_0 ∣+...+∣x_1 −x_0 ∣=  (c^n +c^(n−1) +...+1)∣x_1 −x_0 ∣  c)  x_(n+1) −x_k =(x_(n+1) −x_n )+(x_n −x_(n−1) )+...+(x_(k+1) −x_k )⇒  c  is clear.

$$\left.{a}\right) \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{{n}} \mid\leqslant{c}\mid{x}_{{n}} −{x}_{{n}−\mathrm{1}} \mid\leqslant{c}^{\mathrm{2}} \mid{x}_{{n}−\mathrm{1}} −{x}_{{n}−\mathrm{2}} \mid\leqslant...\leqslant{c}^{{n}} \mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid \\ $$$$\left.{b}\right) \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{\mathrm{0}} \mid=\mid\left({x}_{{n}+\mathrm{1}} −{x}_{{n}} \right)+\left({x}_{{n}} −{x}_{{n}−\mathrm{1}} \right)+...+\left({x}_{\mathrm{1}} −{x}_{\mathrm{0}} \right)\mid\leqslant \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{{n}} \mid+\mid{x}_{{n}} −{x}_{{n}−\mathrm{1}} \mid+...+\mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid\leqslant \\ $$$${c}^{{n}} \mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid+{c}^{{n}−\mathrm{1}} \mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid+...+\mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid= \\ $$$$\left({c}^{{n}} +{c}^{{n}−\mathrm{1}} +...+\mathrm{1}\right)\mid{x}_{\mathrm{1}} −{x}_{\mathrm{0}} \mid \\ $$$$\left.{c}\right) \\ $$$${x}_{{n}+\mathrm{1}} −{x}_{{k}} =\left({x}_{{n}+\mathrm{1}} −{x}_{{n}} \right)+\left({x}_{{n}} −{x}_{{n}−\mathrm{1}} \right)+...+\left({x}_{{k}+\mathrm{1}} −{x}_{{k}} \right)\Rightarrow \\ $$$${c}\:\:{is}\:{clear}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com