Question Number 121154 by zakirullah last updated on 05/Nov/20
Answered by Bird last updated on 05/Nov/20
$${let}\:{f}\left({n}\right)={u}_{{n}} \:\Rightarrow{u}_{{n}} =\mathrm{6}{u}_{{m}−\mathrm{1}} −\mathrm{9}{u}_{{n}−\mathrm{2}} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}\:\:{and}\:{u}_{\mathrm{2}} =\mathrm{2} \\ $$$${e}\Rightarrow{u}_{{n}+\mathrm{2}} =\mathrm{6}{u}_{{n}+\mathrm{1}} −\mathrm{9}{u}_{{n}} \:\Rightarrow \\ $$$${u}_{{n}+\mathrm{2}} −\mathrm{6}{u}_{{n}+\mathrm{1}} +\mathrm{9}{u}_{{n}} =\mathrm{0} \\ $$$$+\left({eq}\right)\rightarrow{r}^{\mathrm{2}} −\mathrm{6}{r}+\mathrm{9}=\mathrm{0}\:\Rightarrow \\ $$$$\left({r}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{r}=\mathrm{3}\left({double}\right)\:\Rightarrow \\ $$$${u}_{{n}} =\left({an}\:+{b}\right)\mathrm{3}^{{n}} \\ $$$${u}_{\mathrm{0}} =\mathrm{1}\:={b} \\ $$$${u}_{\mathrm{2}} =\left(\mathrm{2}{a}+{b}\right)\mathrm{3}^{\mathrm{2}} \:=\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{2}{a}+{b}=\frac{\mathrm{2}}{\mathrm{9}}\:\Rightarrow\mathrm{2}{a}\:=\frac{\mathrm{2}}{\mathrm{9}}−\mathrm{1}\:=\frac{−\mathrm{7}}{\mathrm{9}}\:\Rightarrow \\ $$$${a}=−\frac{\mathrm{7}}{\mathrm{18}}\:\Rightarrow{f}\left({n}\right)=\left(−\frac{\mathrm{7}}{\mathrm{18}}{n}+\mathrm{1}\right)\mathrm{3}^{{n}} \\ $$$$ \\ $$
Commented by zakirullah last updated on 05/Nov/20
$${thsnks}\:{sir}!\:{please}\:{solve}\:{all}? \\ $$
Answered by TANMAY PANACEA last updated on 05/Nov/20
$$\left.\mathrm{2}\right){f}\left({n}\right)={f}\left({n}−\mathrm{1}\right)+\left({n}−\mathrm{1}\right) \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{1}={f}\left(\mathrm{0}\right)+\mathrm{1} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\:\:{f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$$${f}\left({n}\right)=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$
Commented by zakirullah last updated on 05/Nov/20
$${thanks}\:{sir}\:{G} \\ $$
Commented by TANMAY PANACEA last updated on 06/Nov/20
$${most}\:{welcome} \\ $$
Answered by Bird last updated on 05/Nov/20
$${f}\left({n}\right)={f}\left({n}−\mathrm{1}\right)+{n}−\mathrm{1}\:\Rightarrow \\ $$$${u}_{{n}} ={u}_{{n}−\mathrm{1}} +{n}−\mathrm{1}\:\Rightarrow{u}_{{n}+\mathrm{1}} ={u}_{{n}} +{n}\:\Rightarrow \\ $$$${u}_{{n}+\mathrm{1}} −{u}_{{n}} ={n}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left({u}_{{k}+\mathrm{1}} −{u}_{{k}} \right) \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{k}\:=\frac{\left({n}−\mathrm{1}\right){n}}{\mathrm{2}}\:\Rightarrow \\ $$$${u}_{\mathrm{2}} −{u}_{\mathrm{1}} +{u}_{\mathrm{3}} −{u}_{\mathrm{2}} +....+{u}_{{n}} −{u}_{{n}−\mathrm{1}} \\ $$$$=\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\:\Rightarrow{u}_{{n}} −{u}_{\mathrm{1}} =\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$${u}_{{n}} =\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\:+{u}_{\mathrm{1}} \\ $$$${u}_{\mathrm{2}} =\mathrm{1}+{u}_{\mathrm{1}} =\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\mathrm{0}\:\Rightarrow{u}_{{n}} =\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$
Answered by Bird last updated on 05/Nov/20
$${f}\left({n}\right)\:=\mathrm{3}{f}\left({n}−\mathrm{1}\right)\:\Rightarrow{u}_{{n}} =\mathrm{3}{u}_{{n}−\mathrm{1}} \:\Rightarrow \\ $$$$\frac{{u}_{{n}} }{{u}_{{n}−\mathrm{1}} }=\mathrm{3}\:\Rightarrow\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{{u}_{{k}} }{{u}_{{k}−\mathrm{1}} }=\mathrm{3}^{{n}} \:\Rightarrow \\ $$$$\frac{{u}_{\mathrm{1}} }{{u}_{\mathrm{0}} }.\frac{{u}_{\mathrm{2}} }{{u}_{\mathrm{1}} }......\frac{{u}_{{n}} }{{u}_{{n}−\mathrm{1}} }=\mathrm{3}^{{n}} \:\Rightarrow{u}_{{n}} =\mathrm{3}^{{n}} \:{u}_{\mathrm{0}} \\ $$$$\Rightarrow{u}_{{n}} =\mathrm{2}×\mathrm{3}^{{n}} \\ $$
Commented by zakirullah last updated on 05/Nov/20
$${thanks}\:{so}\:{much} \\ $$