Question Number 33154 by Rio Mike last updated on 11/Apr/18
$${it}\:{is}\:{given}\:{that}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{U}_{{n}} =\:\frac{\mathrm{1}+\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{2}×\mathrm{5}^{{n}+\mathrm{1}} }{\mathrm{8}} \\ $$$${where}\:{U}_{{n}} \:{is}\:{the}\:{n}^{{th}} \:{term}\:{of}\:{a}\:{sequence} \\ $$$${find}\:{the}\:{simplified}\:{expression}\:{for}\:{U}_{{n}} \\ $$
Commented by prof Abdo imad last updated on 11/Apr/18
$${we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{u}_{{k}} =\:\frac{\mathrm{1}\:+\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} −\mathrm{2}\:.\:\mathrm{5}^{{n}+\mathrm{1}} }{\mathrm{8}}\:{also} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{u}_{{k}} =\:\frac{\mathrm{1}+\mathrm{3}^{\mathrm{2}{n}} \:−\mathrm{2}\:.\mathrm{5}^{{n}} }{\mathrm{8}}\:\Rightarrow \\ $$$${u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \:+…..+{u}_{{n}} \:−{u}_{\mathrm{1}} \:−{u}_{\mathrm{2}} \:−…−{u}_{{n}−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{3}^{\mathrm{2}{n}+\mathrm{2}} \:−\mathrm{2}.\mathrm{5}^{{n}+\mathrm{1}} \:−\mathrm{1}\:\:−\mathrm{3}^{\mathrm{2}{n}} \:\:+\mathrm{2}.\mathrm{5}^{{n}} }{\mathrm{8}}\:\Rightarrow \\ $$$${u}_{{n}} \:=\:\frac{\mathrm{3}^{\mathrm{2}{n}} \left(\mathrm{9}−\mathrm{1}\right)\:−\mathrm{8}\:.\mathrm{5}^{{n}} }{\mathrm{8}}=\:\mathrm{9}^{{n}} \:−\mathrm{5}^{{n}} \\ $$$${u}_{{n}} =\mathrm{9}^{{n}} \:−\mathrm{5}^{{n}} \:\:. \\ $$$$ \\ $$