已知数列{an}的通项公式是an=2n-1,bn=a1+a2+···+an(n属于n+) 求bn,设cn=【b（n+1）-bn】/3*n求数列{cn}的前n项和

Question

已知数列{an}的通项公式是an=2n-1,bn=a1+a2+···+an(n属于n+) 求bn,设cn=【b（n+1）-bn】/3*n
求数列{cn}的前n项和

Answer 1

因为an=2n-1,bn=a1+a2+···+an,所以bn=n^2,cn=(2n+1)/(3^n),所以Sn=3/(3^1)+5/(3^2)+:::+(2n+1)/(3^n),1/3Sn=3/(3^2)+::::+(2n+1)/(3^(n+1)),两式相减,得2/3Tn=1+2((1/3)^2+(1/3)^3+::::+(1/3)^n)-(2n+1)/(3^(n+1))=4/3-(1/3)^n-(2n+1)/(3^(n+1)),Tn=2-1/2×(1/3)^(n-1)-(2n+1)/(2×(3^n))