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 109834 by 675480065 last updated on 25/Aug/20

Commented by 675480065 last updated on 25/Aug/20

$$\mathrm{Greetings}.$$$$\mathrm{please}\mathrm{i}\mathrm{need}\mathrm{help}$$

Answered by mathmax by abdo last updated on 25/Aug/20

$$1){\mathrm{u}}_0=1,{\mathrm{u}}_1=3\mathrm{and}{\mathrm{u}}_{\mathrm{n}+1}={3\mathrm{u}}_{\mathrm{n}}+{4\mathrm{u}}_{\mathrm{n}-1}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}+2}={3\mathrm{u}}_{\mathrm{n}+1}+{4\mathrm{u}}_{\mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}+2}-{3\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}}=0\to {\mathrm{x}}^2-3\mathrm{x}-4=0$$$$\mathrm{\Delta }=9+16=25\Rightarrow {\mathrm{x}}_1=\frac{3+5}{2}=4\mathrm{and}{\mathrm{x}}_2=\frac{3-5}{2}=-1\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}}=\mathrm{a}{4}^{\mathrm{n}}+\mathrm{b}{(-1)}^{\mathrm{n}}\mathrm{we}\mathrm{have}{\mathrm{u}}_0=\mathrm{a}+\mathrm{b}=1\mathrm{and}{\mathrm{u}}_1=4\mathrm{a}-\mathrm{b}=3\Rightarrow$$$$\{\begin{array}{l}\mathrm{a}+\mathrm{b}=1{}_{\Rightarrow \{\begin{array}{l}\mathrm{b}=1-\mathrm{a}\\ 4\mathrm{a}-1+\mathrm{a}=3\Rightarrow \end{array}}\\ 4\mathrm{a}-\mathrm{b}=3\end{array}$$$$\{\begin{array}{l}5\mathrm{a}=4\\ \mathrm{b}=1-\mathrm{a}\Rightarrow \{\begin{array}{l}\mathrm{a}=\frac{4}{5}\Rightarrow \{\begin{array}{l}\\ \end{array}\\ \mathrm{b}=\frac{1}{5}\end{array}\end{array}$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{4}{5}.4^{\mathrm{n}}+\frac{1}{5}{(-1)}^{\mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}}=\frac{1}{5}\{4^{\mathrm{n}+1}+{(-1)}^{\mathrm{n}}\}$$$$\Rightarrow {\mathrm{u}}_2=\frac{1}{5}\{4^3+1\}=\frac{1}{5}\left\{65\right\}=\frac{65}{5}$$$$$$

Commented by 675480065 last updated on 26/Aug/20

$$\mathrm{thanks}$$

Commented by abdomsup last updated on 26/Aug/20

$$youarewelcome$$

Answered by Aziztisffola last updated on 25/Aug/20

$$\left(\mathrm{i}\right){\mathrm{u}}_2={3\mathrm{u}}_1+{4\mathrm{u}}_0=9+4=13$$$$\left(\mathrm{ii}\right){\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}}=-({\mathrm{u}}_{\mathrm{n}}-{4\mathrm{u}}_{\mathrm{n}-1})$$$$=-(-({\mathrm{u}}_{\mathrm{n}-2}-{4\mathrm{u}}_{\mathrm{n}-3}))$$$$={(-1)}^2({\mathrm{u}}_{\mathrm{n}-2}-{4\mathrm{u}}_{\mathrm{n}-3})$$$$={(-1)}^3({\mathrm{u}}_{\mathrm{n}-4}-{4\mathrm{u}}_{\mathrm{n}-5})$$$$=..............................$$$$={(-1)}^{\mathrm{n}+1}({\mathrm{u}}_1-{4\mathrm{u}}_0)$$$$\left(\mathrm{iii}\right){\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}}={(-1)}^{\mathrm{n}+1}({\mathrm{u}}_1-{4\mathrm{u}}_0)$$$$={(-1)}^{\mathrm{n}+1}(3-4)={(-1)}^{\mathrm{n}+2}$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}}={(-1)}^{\mathrm{n}+2}$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}={(-1)}^{\mathrm{n}+2}+{4\mathrm{u}}_{\mathrm{n}}$$

Answered by mathmax by abdo last updated on 25/Aug/20

$$2)\mathrm{i}\mathrm{think}\mathrm{tbe}\mathrm{relstion}\mathrm{is}{\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}}={(-1)}^{\mathrm{n}}({\mathrm{u}}_1-{4\mathrm{u}}_0)$$$$\mathrm{let}\mathrm{prove}\mathrm{it}\mathrm{by}\mathrm{recurrence}$$$$\mathrm{n}=0\Rightarrow {\mathrm{u}}_1-{4\mathrm{u}}_0={(-1)}^{\mathrm{o}}({\mathrm{u}}_1-{4\mathrm{u}}_0)\mathrm{the}\mathrm{relstion}\mathrm{is}\mathrm{true}$$$$\mathrm{let}\mathrm{suppose}{\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}}={(-1)}^{\mathrm{n}}({\mathrm{u}}_1-{4\mathrm{u}}_0)\mathrm{and}\mathrm{show}$$$${\mathrm{u}}_{\mathrm{n}+2}-{4\mathrm{u}}_{\mathrm{n}+1}={(-1)}^{\mathrm{n}+1}({\mathrm{u}}_1-{4\mathrm{u}}_0)\mathrm{we}\mathrm{have}{\mathrm{u}}_{\mathrm{n}+1}={3\mathrm{u}}_{\mathrm{n}}+{4\mathrm{u}}_{\mathrm{n}-1}\Rightarrow$$$${\mathrm{u}}_{\mathrm{n}+2}={3\mathrm{u}}_{\mathrm{n}+1}+{4\mathrm{u}}_{\mathrm{n}}\Rightarrow {\mathrm{u}}_{\mathrm{n}+2}-{4\mathrm{u}}_{\mathrm{n}+1}=-{\mathrm{u}}_{\mathrm{n}+1}+{4\mathrm{u}}_{\mathrm{n}}$$$$=-({\mathrm{u}}_{\mathrm{n}+1}-{4\mathrm{u}}_{\mathrm{n}})=-{(-1)}^{\mathrm{n}}({\mathrm{u}}_1-{4\mathrm{u}}_0)={(-1)}^{\mathrm{n}+1}({\mathrm{u}}_1-{4\mathrm{u}}_0)$$$$\mathrm{the}\mathrm{relation}\mathrm{is}\mathrm{true}\mathrm{st}\mathrm{term}\mathrm{n}+1$$$$$$

Commented by 675480065 last updated on 26/Aug/20

$$\mathrm{thanks}\mathrm{so}\mathrm{much}.$$$$\mathrm{is}\mathrm{it}\mathrm{possible}\mathrm{if}\mathrm{you}\mathrm{can}\mathrm{complete}\mathrm{it}.?\mathrm{pls}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com