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Question Number 119250 by Pengu last updated on 23/Oct/20

Commented by Pengu last updated on 23/Oct/20

$$\mathrm{I}\mathrm{know}{a}_n=a_n^h+a_n^p$$$$a_n^h\Rightarrow a_n=2a_{n-1}$$$$r^n=a^n\Rightarrow r=2$$$$\therefore a_n^h=\alpha 2^{n-1}$$$$$$$$\mathrm{I}\mathrm{do}\mathrm{not}\mathrm{understand}\mathrm{how}\mathrm{to}\mathrm{solve}{a}_n^p$$

Commented by Pengu last updated on 23/Oct/20

Also, hello. I am not new. I am Filup, and have used this app for many years, but forgot my various account details. I wonder if anyone remembers me �� Filup, FilupErwinSmith, and Penguin have been my previous aliases. I am finally doing a mathematics degree, after many years.

Commented by mr W last updated on 23/Oct/20

$$welcomebacksir!$$$$istillrememberthediscussionabout$$$$theinterestingquestion12883from$$$$yousomeyearsago.$$

Commented by Pengu last updated on 23/Oct/20

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{very}\mathrm{glad}\mathrm{to}\mathrm{see}\mathrm{you}\mathrm{remember}!$$

Commented by prakash jain last updated on 24/Oct/20

Hi Filup, How are you?

Answered by 1549442205PVT last updated on 23/Oct/20

$$\mathrm{a})\mathrm{Given}{\mathrm{a}}_{\mathrm{n}}={2\mathrm{a}}_{\mathrm{n}-1}+2^{\mathrm{n}}$$$$\mathrm{Consider}\mathrm{solution}{\mathrm{a}}_{\mathrm{n}}={\mathrm{n}2}^{\mathrm{n}}\Rightarrow {\mathrm{a}}_{\mathrm{n}-1}=(\mathrm{n}-1)2^{\mathrm{n}-1}$$$$\Rightarrow {2\mathrm{a}}_{\mathrm{n}-1}=(\mathrm{n}-1).2^{\mathrm{n}}\Rightarrow {2\mathrm{a}}_{\mathrm{n}-1}+2^{\mathrm{n}}=\mathrm{n}.2^{\mathrm{n}}$$$$\Rightarrow {\mathrm{a}}_{\mathrm{n}}={2\mathrm{a}}_{\mathrm{n}-1}+2^{\mathrm{n}}\mathrm{which}\mathrm{shows}\mathrm{that}$$$${\mathrm{a}}_{\mathrm{n}}={\mathrm{n}2}^{\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{concurrent}$$$$\mathrm{relation}$$$$\mathrm{b})\mathrm{Put}{\mathrm{a}}_{\mathrm{n}}=2^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}\Rightarrow {\mathrm{a}}_{\mathrm{n}+1}={2\mathrm{a}}_{\mathrm{n}}+2^{\mathrm{n}+1}$$$$\iff 2^{\mathrm{n}+1}{\mathrm{u}}_{\mathrm{n}+1}=2.2^{\mathrm{n}}{\mathrm{u}}_{\mathrm{n}}+2^{\mathrm{n}+1}$$$$\Rightarrow {\mathrm{u}}_{\mathrm{n}+1}={\mathrm{u}}_{\mathrm{n}}+1\Rightarrow \left({\mathrm{u}}_{\mathrm{n}}\right)\mathrm{is}\mathrm{a}\mathrm{A}.\mathrm{P}.\mathrm{with}$$$$\mathrm{first}\mathrm{term}\mathrm{equal}\mathrm{to}{\mathrm{u}}_0={\mathrm{a}}_0\mathrm{and}$$$$\mathrm{common}\mathrm{difference}\mathrm{equal}\mathrm{to}1.\mathrm{We}\mathrm{infer}$$$${\mathrm{u}}_{\mathrm{n}}={\mathrm{a}}_0+\mathrm{n}\Rightarrow {\mathrm{a}}_{\mathrm{n}}=({\mathrm{a}}_0+\mathrm{n})2^{\mathrm{n}}.\mathrm{Thus},$$$${\mathrm{a}}_{\mathrm{n}}=(\mathrm{n}+{\mathrm{a}}_0)2^{\mathrm{n}}\mathrm{is}\mathrm{general}\mathrm{term}\mathrm{we}\mathrm{need}\mathrm{find}$$$$\mathrm{c})\mathrm{For}{\mathrm{a}}_0=2\mathrm{we}\mathrm{get}{\mathrm{a}}_{\mathrm{n}}=(\mathrm{n}+2)2^{\mathrm{n}}$$

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