Question

数列an 的前n项和为Sn 且满足3an=2sn-4n+9(1)求an的通项公式(2)设bn=(

Answer 1

由题意 3a1 = 2a1-4+9

所以 a1 = 5

2Sn = 3an+4n-9

2S(n-1) = 3a(n-1)+4(n-1)-9

所以 2Sn-2S(n-1) = 3[an-a(n-1)]+4 = 2an

所以 an = 3a(n-1) -4

所以 an -2 = 3a(n-1) -6 = 3[a(n-1) -2]

所以 数列{an -2}是首项为3,公比为3的等比数列,即 an -2 = 3^n

所以 an = 3^n +2

bn = (2n-1)an = (2n-1)*(3^n +2) = (2n-1)*3^n +(4n-2)

所以 Tn = 3^1 +2 +3*3^2 +6 + 5*3^3 +10 +7*3^4 +14 +...+(2n-1)*3^n +(4n-2)

所以 Tn = 3^1 +3*3^2 +5*3^3 +7*3^4 +...+(2n-1)*3^n +2+6+10+14+...+(4n-2)

所以 Tn = 3^1 +3*3^2 +5*3^3 +7*3^4 +...+(2n-1)*3^n +2n^2

所以 3Tn = 3^2 +3*3^3 +5*3^4 +7*3^4 +...+(2n-1)*3^(n+1) +6n^2

所以 2Tn = 3Tn-Tn = -3 -2*3^2 -2*3^3 -2*3^4 -...-2*3^n +(2n-1)*3^(n+1) +4n^2

所以 2Tn = -3 -18[3^(n-1) -1]/(3-1) +(2n-1)*3^(n+1) +4n^2

所以 2Tn = -3 -3^(n+1) +9 +(2n-1)*3^(n+1) +4n^2

所以 2Tn = (2n-2)*3^(n+1) +4n^2 +6

所以 Tn = (n-1)*3^(n+1) +2n^2 +3