数列(an)中满足a1=2,a2=1且an+2an-1-3an-2=0(n≥3),1则an=
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数列(an)中满足a1=2,a2=1且an+2an-1-3an-2=0(n≥3),1则an=
数列(an)中满足a1=2,a2=1且an+2an-1-3an-2=0(n≥3),1则an=
数列(an)中满足a1=2,a2=1且an+2an-1-3an-2=0(n≥3),1则an=
an+2a(n-1)-3a(n-2)=0(n≥3),
所以an -a(n-1)+3a(n-1)-3a(n-2)=0
即an -a(n-1) =-3 [a (n-1)-a(n-2)]
所以数列{ an -a(n-1)}是等比数列,首项为a2-a1=-1,公比为-3.
an -a(n-1)=-(-3)^(n-1),
∴an=a1+(a2-a1)+(a3-a2)+……+( an -a(n-1))
=2-(-3)^0-(-3)^1-(-3)^2-……-(-3)^(n-1)
=2-[1-(-3)^(n-1)]/[1-(-3)]
=2-[1-(-3)^(n-1)]/4
=[7+(-3)^(n-1)]/4.
an+2a(n-1)-3a(n-2)=0
即 an= -2a(n-1)+3a(n-2)
an -a(n-1)= -3a(n-1)+3a(n-2)=-3[a(n-1) -a(n-2)]
[an -a(n-1)] / [a(n-1) -a(n-2)]=-3
即数列为后项与前项之差的等比数列(二级等比数列),公比q=-3
a1 2 ...
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an+2a(n-1)-3a(n-2)=0
即 an= -2a(n-1)+3a(n-2)
an -a(n-1)= -3a(n-1)+3a(n-2)=-3[a(n-1) -a(n-2)]
[an -a(n-1)] / [a(n-1) -a(n-2)]=-3
即数列为后项与前项之差的等比数列(二级等比数列),公比q=-3
a1 2
a2 1 -1
a3 4 3 4=1+3=1-(-3)^1
a4 -5 -9 -5=4-9=4-(-3)^2=1-(-3)^1 -(-3)^2
a5 22 27 22=-5+27=-5-(-3)^3=1-(-3)^1 -(-3)^2 -(-3)^3
a6 -59 -81 -59=22-81=22 -(-3)^4=1-(-3)^1 -(-3)^2 -(-3)^3 -(-3)^4
......
an=1-(-3)^1 -(-3)^2 -(-3)^3 -(-3)^4 -......-(-3)^(n-2) (n≥3) ]
=1-[(-3)^1 +(-3)^2 +(-3)^3 +(-3)^4 +......+(-3)^(n-2)]
=1+[(-3)^(n-1) +3]/4
=(-3)^(n-1) /4 + 7/4
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