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a-1-1-a-2-2-a-n-1-n-1-a-n-2a-n-1-find-a-n-




Question Number 81910 by mr W last updated on 16/Feb/20
a_1 =1  a_2 =2  a_(n+1) =(n+1)a_n −2a_(n−1)   find a_n =?
$${a}_{\mathrm{1}} =\mathrm{1} \\ $$$${a}_{\mathrm{2}} =\mathrm{2} \\ $$$${a}_{{n}+\mathrm{1}} =\left({n}+\mathrm{1}\right){a}_{{n}} −\mathrm{2}{a}_{{n}−\mathrm{1}} \\ $$$${find}\:{a}_{{n}} =? \\ $$
Commented by jagoll last updated on 16/Feb/20
a_3  = 3a_2 −2a_1  = 6−2=4  a_(4 ) = 4a_3 −2a_2  = 16−4=12  a_5  = 5a_4 −2a_3  = 60 − 8=52  a_6  = 6a_5  − 2a_4  = 312−24=288
$${a}_{\mathrm{3}} \:=\:\mathrm{3}{a}_{\mathrm{2}} −\mathrm{2}{a}_{\mathrm{1}} \:=\:\mathrm{6}−\mathrm{2}=\mathrm{4} \\ $$$${a}_{\mathrm{4}\:} =\:\mathrm{4}{a}_{\mathrm{3}} −\mathrm{2}{a}_{\mathrm{2}} \:=\:\mathrm{16}−\mathrm{4}=\mathrm{12} \\ $$$${a}_{\mathrm{5}} \:=\:\mathrm{5}{a}_{\mathrm{4}} −\mathrm{2}{a}_{\mathrm{3}} \:=\:\mathrm{60}\:−\:\mathrm{8}=\mathrm{52} \\ $$$${a}_{\mathrm{6}} \:=\:\mathrm{6}{a}_{\mathrm{5}} \:−\:\mathrm{2}{a}_{\mathrm{4}} \:=\:\mathrm{312}−\mathrm{24}=\mathrm{288} \\ $$

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