an=3n-1,bn=2^n,设Tn=anb1+a(n-1)b2+……+a1bn(n∈N+),证明:Tn+12=-2an+10bn(n∈N+)
请大侠 看看我的问题...
Tn=2an+2²a(n-1)+2∧3a(n-2)+…+2∧na1①
2Tn=2²an+2∧3a(n-1)+2∧4a(n-2)+…2∧(n+1)a1②
②-①Tn=2(a1+.+an) 但是我做错了?为什么 请大侠说说,顺便给我正确的.我会追问道.
人气:354 ℃ 时间:2020-09-01 13:54:12
解答
应该消掉a1,.,an,留下2^2,.,2^n等
Tn=2an+2²a(n-1)+2∧3a(n-2)+…+2∧na1①
2Tn=2²an+2∧3a(n-1)+2∧4a(n-2)+…2∧(n+1)a1②
错位相减法an=3n-1即an-a(n-1)=a(n-1)-a(n-2)=.=a2-a1=3
②-①Tn=-2an+2²[an-a(n-1)]+2∧3[a(n-1)-a(n-2)]+…+2∧n[a2-a1]+2∧(n+1)a1
=-2an+3*2²+3*2∧3+…+3*2∧n+2∧(n+1)*2
=-2an+3(2²+2∧3+…+2∧n)+2∧(n+2)
=-2an+3[-4+2^(n+1)]+2∧(n+2)
=-2(3n-1)-12+3*2^(n+1)+2*2∧(n+1)
=5*2^(n+1)-6n+2-12
左边=Tn+12=5*2^(n+1)-6n+2-12+12=5*2^(n+1)-6n+2
右边=-2an+10bn
=-2(3n-1)+10*2^n
=-2(3n-1)+5*2*2^n
=-6n+2+5*2^(n+1)
即证
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