数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn
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数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn数列前n项和为s
数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn
数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn
数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn
证明:由题意:an+Sn=2n……(1),所以a(n+1)+S(n+1)=2(n+1)……(2)
用(2)-(1)得:2a(n+1)-an=2,即2[a(n+1)-2]=an-2,即[a(n+1)-2]/[an-2]=1/2——为常数
所以{an-2}是首项为-1公比是1/2的等比数列
an-2=(-1)(1/2)^(n-1),所以Sn=2n-an=2n-2+(1/2)^(n-1)
设Cn=an+Sn,则C1=a1+S1=2a1=2,则Cn=2n=an+Sn [1] ,则C
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