Question

已知函数f(x)=1/2x^2+3/2x,数列an的前n项和为Sn,点(n,Sn)[n属于N*]均在函数y=f(x)

1.求数列an的通项公式an
2.令bn=an/2^(n-1)，求数列bn的前n项和Tn
3.令cn=an/a(n+1)+a(n+1)/an,证明：2n<c1+c2+……+cn<2n+1/2

麻烦给出步骤，谢谢~
如果第三问太复杂，请帮忙解答一下前两问，thx

Answer 1

1.
Sn=(1/2)n^2+(3/2)n
n>=2时，
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时，A1=S1=1/2+3/2=2也满足上式

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)
两边同乘2
2Tn=4+3/2^0+4/2^1+……+(n+1)/2^(n-2)
两式错位相减
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)<1/2
C1+C2+……+Cn-2n<1/2
C1+C2+……+Cn<2n+1/2

设a=(n+1)/(n+2) b=(n+2)/(n+1) 则ab=1
a+b>=2(ab)^0.5 当且仅当a=b时，不等式取等号
Cn=a+b>=2(ab)^0.5=2
若a=b 则(n+1)/(n+2)=(n+2)/(n+1) 解得n=-3/2，不属于N
不等式等号不成立
Cn>2
C1+C2+……+Cn>2n

综上所述
2n<C1+C2+……+Cn<2n+1/2

Answer 2

Sn=1/2n^2+3/2n
a1=S1=1/2*1^2+3/2*1
=2
an=Sn-S<n-1>
下面你先做

Answer 3

没有分 谁愿意做？