作业帮已知函数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an2.令bn=an/2^(n-1),求数列bn的前n项和Tn3.令cn=an/a(n+1)+a(n+1)/an,证明：2n
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![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an2.令bn=an/2^(n-1),求数列bn的前n项和Tn3.令cn=an/a(n+1)+a(n+1)/an,证明：2n](/uploads/image/z/2337561-9-1.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D1%2F2x%5E2%2B3%2F2x%2C%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E7%82%B9%28n%2CSn%29%5Bn%E5%B1%9E%E4%BA%8EN%2A%5D%E5%9D%87%E5%9C%A8%E5%87%BD%E6%95%B0y%3Df%28x%291.%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8Fan2.%E4%BB%A4bn%3Dan%2F2%5E%28n-1%29%2C%E6%B1%82%E6%95%B0%E5%88%97bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn3.%E4%BB%A4cn%3Dan%2Fa%28n%2B1%29%2Ba%28n%2B1%29%2Fan%2C%E8%AF%81%E6%98%8E%EF%BC%9A2n)
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an2.令bn=an/2^(n-1),求数列bn的前n项和Tn3.令cn=an/a(n+1)+a(n+1)/an,证明：2n
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
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an2.令bn=an/2^(n-1),求数列bn的前n项和Tn3.令cn=an/a(n+1)+a(n+1)/an,证明：2n
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)2
C1+C2+……+Cn>2n
综上所述
2n
Sn=1/2n^2+3/2n
a1=S1=1/2*1^2+3/2*1
=2
an=Sn-S
下面你先做
没有分 谁愿意做？