(n,Sn)都在函數(shù)f(x)=2x^2-x上,
即Sn=2n^2-n
an+1=Sn+1-Sn=[2(n+1)^2-(n+1)]-(2n^2-n)=4n+1
an=4(n-1)+1
cn=2/[(4(n-1)+1)(4n+1)]=1/2{1/[(4(n-1)+1)]-1/[(4n+1)]}
Tn=1/2{1 - 1/[(4n+1)]}
=2n/(4n+1)
當(dāng)m=10時
Tn=2/(4+1/n)<2/4=10/20=m/20
而當(dāng)m=9時
T2=2/(4+1/2)=4/9>9/20=m/20
所以最小正整數(shù)m=10當(dāng)m=10時Tn=2/(4+1/n)<2/4=10/20=m/20而當(dāng)m=9時 T2=2/(4+1/2)=4/9>9/20=m/20所以最小正整數(shù)m=10 這步怎么求出的m=10m/20=1/2 Tn=2n/(4n+1)=2/(4+1/n)<2/(4+0)=1/2 Tn10時也有Tn18/38=9/19T2>m/20就是說m至少取10能保證對任何正整數(shù)n都滿足Tn40/(4+1/n)而n越大40/(4+1/n)越大，n趨于無窮大時，40/(4+1/n) 趨于10，但始終有40/(4+1/n)<10因此m>=10時?？傆蠺n