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Question Number 131158 by EDWIN88 last updated on 02/Feb/21
If 4a_n +2a_(−n) =3n^2 +2n−3 find a_n =?
$${If}\:\mathrm{4}{a}_{{n}} +\mathrm{2}{a}_{−{n}} =\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{3} \\ $$$${find}\:{a}_{{n}} \:=? \\ $$
Answered by mr W last updated on 02/Feb/21
4a_n +2a_(−n) =3n^2 +2n−3 4a_(−n) +2a_n =3n^2 −2n−3 8a_n +4a_(−n) =6n^2 +4n−6 ⇒6a_n =3n^2 +6n−3 ⇒a_n =(1/2)(n^2 +2n−1)
$$\mathrm{4}{a}_{{n}} +\mathrm{2}{a}_{−{n}} =\mathrm{3}{n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{3} \\ $$$$\mathrm{4}{a}_{−{n}} +\mathrm{2}{a}_{{n}} =\mathrm{3}{n}^{\mathrm{2}} −\mathrm{2}{n}−\mathrm{3} \\ $$$$\mathrm{8}{a}_{{n}} +\mathrm{4}{a}_{−{n}} =\mathrm{6}{n}^{\mathrm{2}} +\mathrm{4}{n}−\mathrm{6} \\ $$$$\Rightarrow\mathrm{6}{a}_{{n}} =\mathrm{3}{n}^{\mathrm{2}} +\mathrm{6}{n}−\mathrm{3} \\ $$$$\Rightarrow{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left({n}^{\mathrm{2}} +\mathrm{2}{n}−\mathrm{1}\right) \\ $$

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