Senior three a round of sequence sequence sum dislocation subtraction homework master into ah It is known that the tolerance D of the arithmetic sequence an is greater than 0, and A2 and A5 are two of the equations x ^ 2-12x + 27 = 0. The sum of the first n terms of the sequence BN is TN, and TN = 1-1 / 2bn (1) Finding {an} {BN} will find that an = 2N-1, BN = 2 / 3 * [(1 / 3) of the N-1 power] (2) Note CN = anbn, find the first n terms of {CN} and Sn Is to seek the second question of dislocation subtraction!

Senior three a round of sequence sequence sum dislocation subtraction homework master into ah It is known that the tolerance D of the arithmetic sequence an is greater than 0, and A2 and A5 are two of the equations x ^ 2-12x + 27 = 0. The sum of the first n terms of the sequence BN is TN, and TN = 1-1 / 2bn (1) Finding {an} {BN} will find that an = 2N-1, BN = 2 / 3 * [(1 / 3) of the N-1 power] (2) Note CN = anbn, find the first n terms of {CN} and Sn Is to seek the second question of dislocation subtraction!

an=2n-1,bn=2/3*[(1/3)^(n-1)]
∵cn=anbn
∴cn=2/3*(2n-1)(1/3)^(n-1)
∴Sn=2/3(1*1+3*1/3+5*(1/3)^2+…… +(2n-3)*(1/3)^(n-2)+(2n-1)(1/3)^(n-1))
1/3Sn=2/3(1*1/3+3*(1/3)^2+5(1/3)^3+…… +(2n-3)(1/3)^(n-1)+(2n-1)*(1/3)^n)
The two formulas are subtracted to get the
2/3Sn=2/3(1+2*1/3+2*(1/3)^2+…… +2*(1/3)^(n-1)-(2n-1)*(1/3)^n)
∴Sn=2(1+1/3+(1/3)^2+…… +(1/3)^(n-1))-(2n-1)(1/3)^n-1
=3(1-(1/3)^n)-(2n-1)(1/3)^n-1
=2-(2n-1+3)(1/3)^n
=2-2(n+1)(1/3)^n