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)上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=1/a(n-1),证明:c2^2+c3^2+c4^2+…+cn^2
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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)上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=1/a(n-1),证明:c2^2+c3^2+c4^2+…+cn^2](/uploads/image/z/3501807-15-7.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%29%E4%B8%8A1.%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%3D1%2Fa%28n-1%29%2C%E8%AF%81%E6%98%8E%EF%BC%9Ac2%5E2%2Bc3%5E2%2Bc4%5E2%2B%E2%80%A6%2Bcn%5E2)
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)上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=1/a(n-1),证明:c2^2+c3^2+c4^2+…+cn^2
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)上
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Tn3.令cn=1/a(n-1),证明:c2^2+c3^2+c4^2+…+cn^2
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)上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=1/a(n-1),证明:c2^2+c3^2+c4^2+…+cn^2
1,Sn=f(n)=1/2*n^2+3/2*n
当n=1时,a1=S1=1/2+3/2=2;
当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
而当n=1时,a1=2=1+1,满足此时
所以an=n+1 (n∈N+)
2,bn=(n+1)*(1/2)^(n-1)
Tn=2*(1/2)^0+3*(1/2)^1+……+(n+1)*(1/2)^(n-1)
那么1/2*Tn=2*(1/2)^1+……+n*(1/2)^(n-1)+(n+1)*(1/2)^n
两式相减,得:1/2*Tn=2+(1/2)^1+……+(1/2)^(n-1)-(n+1)*(1/2)^n
=2+(1/2)*[1-(1/2)^(n-1)]/(1-1/2)-(n+1)*(1/2)^n
=3-(n+3)*(1/2)^n
所以Tn=6-(n+3)*(1/2)^(n-1)
3,cn=1/n,那么当n>1时,cn^2=1/n^2
Sn=f(n)点(n,Sn)[n属于N*]均在函数y=f(x)上
Sn=1/2n^2+3/2n
Sn-1=1/2(n-1)^2+3/2(x-1)
an=Sn-s(n-1)=1/2(n+n-1)(n-n+1)+3/2(x-x+1)=1/2(2n-1)+3/2=n+1
bn=an/2^(n-1)=(n+1)/2^(n-1)=n/2^(n-1)+1/2^(n-1)
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Sn=f(n)点(n,Sn)[n属于N*]均在函数y=f(x)上
Sn=1/2n^2+3/2n
Sn-1=1/2(n-1)^2+3/2(x-1)
an=Sn-s(n-1)=1/2(n+n-1)(n-n+1)+3/2(x-x+1)=1/2(2n-1)+3/2=n+1
bn=an/2^(n-1)=(n+1)/2^(n-1)=n/2^(n-1)+1/2^(n-1)
Cn=n/2^(n-1) Dn=1/2^(n-1)
Scn=1/1+2/2+3/2^2+4/2^3+5/2^4+...+(n-1)/2^(n-2)+(n)/2^(n-1)
1/2Scn= 1/2+2/2^2+3/2^3+4/2^4+...+(n-2)/2^(n-2)+(n-1)/2^(n-1)+n/2^n
Scn-1/2Scn=1+1/2+1/2^2+1/2^3+1/2^4+...+1/2^(n-2)+1/2^(n-1)-n/2^n
1/2Scn=1*(1-(1/2)^n)/(1-1/2)-n/2^n=2-2^(1-n)-n*2^(-n)
Scn=4-2^(2-n)-n*2^(-n)
Sdn=1+1/2+1/4+1/2^3+...+1/2^(n-1)=2-2^(1-n)
Tn=Scn+Sdn=6-(4+2)2^(-n)-n2^(-n)=6-(6+n)/2^n
cn=1/a(n-1)=1/(n-1+1)=1/n
c2^2+c3^2+c4^2+…+cn^2
=1/2^2+1/3^2+1/4^2+...+1/n^2<1
证明:
1/n^2<1/(n(n-1))=1/(n-1)-1/n
(1/2^2+1/3^2+1/4^2+...+1/n^2)<1-1/2+1/2-1/3+1/3-1/4+...+1/(n-2)-1/(n-1)+1/(n-1)-1/n=1-1/n<1
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