1.
Sn=(1/2)n^2+(3/2)n
n>=2時,
An=Sn-S(n-1)
=(1/2)n^2+(3/2)n-(1/2)(n-1)^2-(3/2)(n-1)
=n+1
n=1時,A1=S1=1/2+3/2=2也滿足上式
2.
Bn=An/2^(n-1)=(n+1)/2^(n-1)
Tn=B1+B2+B3+……+Bn
=2/2^0+3/2^1+4/2^2+……+(n+1)/2^(n-1)
兩邊同乘2
2Tn=4+3/2^0+4/2^1+……+(n+1)/2^(n-2)
兩式錯位相減
2Tn-Tn=4+[(3/1-2/1)+(4/2-3/2)+……+(n+1)/2^(n-2)-n/2^(n-2)]-(n+1)/2^(n-1)
=4+(1+1/2+……+1/2^(n-2)-(n+1)/2^(n-1)
=4+(1-1/2^(n-1))/(1-1/2)-(n+1)/2^(n-1)
=6-(n+3)/2^(n-1)
3.
Cn=An/A(n+1)+A(n+1)/An=(n+1)/(n+2)+(n+2)/(n+1)
Cn-2=(n+1)/(n+2)+(n+2)/(n+1)-2
=(n+2-1)/(n+2)+(n+1+1)/(n+1)-2
=1-1/(n+2)+1+1/(n+1)-2
=1/(n+1)-1/(n+2)
C1+C2+……+Cn-2n
=(1/2-1/3)+(1/3-1/4)+……+1/(n+1)-1/(n+2)
=1/2-1/(n+2)2
C1+C2+……+Cn>2n
綜上所述
2n