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Question Number 98380 by Rio Michael last updated on 13/Jun/20

$$\mathrm{let}\left\{u_n\right\}\mathrm{and}\left\{v_n\right\}\mathrm{be}\mathrm{sequences}\mathrm{defined}\mathrm{by}$$$$u_0=9,u_{n+1}=\frac{1}{2}{u}_n-3.$$$$v_n=u_n+6.$$$$\mathrm{Calculate}{P}_n=\underset{i=0}{\sum ^n}{V}_i\mathrm{in}\mathrm{terms}\mathrm{of}n,\mathrm{the}\mathrm{deduce}{Q}_n=\underset{i=0}{\sum ^n}{u}_i$$$$\mathrm{using}\mathrm{the}\mathrm{above}\mathrm{expressions}\mathrm{find}$$$$\underset{x\to \mathrm{\infty }}{\lim }{Q}_n.$$

Answered by Aziztisffola last updated on 13/Jun/20

$${\mathrm{P}}_{\mathrm{n}}=30-15{\left(\frac{1}{2}\right)}^{\mathrm{n}}$$$${\mathrm{Q}}_{\mathrm{n}}=2(1-{\left(\frac{1}{2}\right)}^{\mathrm{n}+1})-6(\mathrm{n}+1)$$$$\text{Double subscripts: use braces to clarify}$$

Answered by mr W last updated on 13/Jun/20

$$u_{n+1}=\frac{1}{2}{u}_n-3$$$$u_{n+1}+6=\frac{1}{2}(u_n+6)$$$$\Rightarrow u_n+6=(u_0+6){\left(\frac{1}{2}\right)}^n=\frac{15}{2^n}$$$$\Rightarrow u_n=\frac{15}{2^n}-6$$$$\Rightarrow v_n=\frac{15}{2^n}$$$$P_n=\underset{i=0}{\sum ^n}{v}_i=15\times \frac{1-\frac{1}{2^{n+1}}}{1-\frac{1}{2}}=30(1-\frac{1}{2^{n+1}})$$$$Q_n=\underset{i=0}{\sum ^n}{u}_i=\underset{i=0}{\sum ^n}(v_i-6)=30(1-\frac{1}{2^{n+1}})-6(n+1)$$$$\underset{n\to \mathrm{\infty }}{\lim }{Q}_n=-\mathrm{\infty }$$

Commented by Rio Michael last updated on 13/Jun/20

$$\mathrm{perfect}\mathrm{now}\mathrm{sir}\mathrm{thanks}$$

Commented by Rio Michael last updated on 13/Jun/20

$$\mathrm{sir}\mathrm{that}\mathrm{is}\mathrm{correct}.\mathrm{thanks}\mathrm{to}\mathrm{you}\mathrm{both}.$$$$\mathrm{my}\mathrm{worry}\mathrm{is}:\mathrm{why}(n+1)?$$$$\mathrm{i}\mathrm{know}\mathrm{\Sigma }{V}_i=\mathrm{\Sigma }{u}_i+\mathrm{\Sigma }6\Rightarrow \mathrm{\Sigma }{u}_i=\mathrm{\Sigma }{V}_i-\mathrm{\Sigma }6$$$$\mathrm{why}\mathrm{will}\mathrm{it}\mathrm{not}\mathrm{equal}30[1-{\left(\frac{1}{2}\right)}^n]-6n\mathrm{instead}\mathrm{it}\mathrm{equals}$$$$30[1-{\left(\frac{1}{2}\right)}^{n+1}]-6n-6$$

Commented by mr W last updated on 13/Jun/20

$$becausei=0ton,totallyn+1times,$$$$and{P}_0\ne 0,Q_0\ne 0.$$$$\underset{i=0}{\sum ^n}{P}_i\ne \underset{i=1}{\sum ^n}{P}_i$$$$\underset{i=0}{\sum ^n}{Q}_i\ne \underset{i=1}{\sum ^n}{Q}_i$$

Answered by mathmax by abdo last updated on 13/Jun/20

$${\mathrm{v}}_{\mathrm{n}}={\mathrm{u}}_{\mathrm{n}}+6\Rightarrow {\mathrm{v}}_{\mathrm{n}+1}={\mathrm{u}}_{\mathrm{n}+1}+6=\frac{1}{2}{\mathrm{u}}_{\mathrm{n}}-3+6=\frac{1}{2}{\mathrm{u}}_{\mathrm{n}}+3=\frac{1}{2}({\mathrm{u}}_{\mathrm{n}}+6)=\frac{1}{2}{\mathrm{v}}_{\mathrm{n}}$$$$\Rightarrow {\mathrm{v}}_{\mathrm{n}}\mathrm{is}\mathrm{g}.\mathrm{prog}.\Rightarrow {\mathrm{v}}_{\mathrm{n}}={\mathrm{v}}_0{\left(\frac{1}{2}\right)}^{\mathrm{n}}\mathrm{we}\mathrm{have}{\mathrm{v}}_0={\mathrm{u}}_0+6=15\Rightarrow$$$${\mathrm{v}}_{\mathrm{n}}=\frac{15}{2^{\mathrm{n}}}\mathrm{we}\mathrm{have}{\mathrm{P}}_{\mathrm{n}}=\sum _{\mathrm{i}=0}^{\mathrm{n}}{\mathrm{v}}_{\mathrm{i}}={\mathrm{v}}_0\times \frac{1-{\mathrm{q}}^{\mathrm{n}+1}}{1-\mathrm{q}}=15\times \frac{1-\frac{1}{2^{\mathrm{n}+1}}}{\frac{1}{2}}$$$$=15(2-\frac{2}{2^{\mathrm{n}+1}})=15(2-\frac{1}{2^{\mathrm{n}}})\Rightarrow {\mathrm{P}}_{\mathrm{n}}=30-\frac{15}{2^{\mathrm{n}}}$$$${\mathrm{Q}}_{\mathrm{n}}=\sum _{\mathrm{i}=0}^{\mathrm{n}}{\mathrm{u}}_{\mathrm{i}}=\sum _{\mathrm{i}=0}^{\mathrm{n}}({\mathrm{v}}_{\mathrm{i}}-6)=\sum _{\mathrm{i}=0}^{\mathrm{n}}{\mathrm{v}}_{\mathrm{i}}-6(\mathrm{n}+1)$$$$=30-\frac{15}{2^{\mathrm{n}}}-6(\mathrm{n}+1)\Rightarrow {\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{Q}}_{\mathrm{n}}=-\mathrm{\infty }$$

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