数列 谢已知数列 {an} 满足a1=1,当n为奇数,an+1=(1/2)*an+n;当n为偶数,an+1=an-2n,
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数列 谢
已知数列 {an} 满足a1=1,当n为奇数,an+1=(1/2)*an+n;当n为偶数,an+1=an-2n,1)设bn=a2n-2(n∈N*),求证:{bn}是等比数列,并求bn
已知数列 {an} 满足a1=1,当n为奇数,an+1=(1/2)*an+n;当n为偶数,an+1=an-2n,1)设bn=a2n-2(n∈N*),求证:{bn}是等比数列,并求bn
∵bn=a(2n)-2∴b(n+1)=a(2n+2)-2
∴b(n+1)/bn=[a(2n+2)-2]/[a(2n)-2]
=[(1/2)a(2n+1)+2n-1]/[a(2n)-2]
={(1/2)[a(2n)-4n]+2n-1}/[a(2n)-2]
=[(1/2)a(2n)-1]/[a(2n)-2]
=(1/2)[a(2n)-2]/[a(2n)-2]
=1/2
∵1/2是与n无关常数
∴{bn}是等比数列,q=1/2
b1=3/2,bn=3*(1/2)^n
∴b(n+1)/bn=[a(2n+2)-2]/[a(2n)-2]
=[(1/2)a(2n+1)+2n-1]/[a(2n)-2]
={(1/2)[a(2n)-4n]+2n-1}/[a(2n)-2]
=[(1/2)a(2n)-1]/[a(2n)-2]
=(1/2)[a(2n)-2]/[a(2n)-2]
=1/2
∵1/2是与n无关常数
∴{bn}是等比数列,q=1/2
b1=3/2,bn=3*(1/2)^n
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