设数列{an}的前n项和为Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n (n∈N*).
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设数列{an}的前n项和为Sn,已知a1+2a2+3a3+…+nan=(n-1)Sn+2n (n∈N*).
(1)求a2,a3的值;
(2)求证:数列{Sn+2}是等比数列.
(1)求a2,a3的值;
(2)求证:数列{Sn+2}是等比数列.
(1)∵a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*),
∴当n=1时,a1=2×1=2; (2分)
当n=2时,a1+2a2=(a1+a2)+4,∴a2=4; (5分)
当n=3时,a1+2a2+3a3=2(a1+a2+a3)+6,∴a3=8.(8分)
(2)证明:∵a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*),①
∴当n≥2时,a1+2a2+3a3+…+(n-1)an-1=(n-2)Sn-1+2(n-1).②(9分)
①-②得nan=(n-1)Sn-(n-2)Sn-1+2
∴nan=n(Sn-Sn-1)-Sn+2Sn-1+2
∴nan=nan-Sn+2Sn-1+2.(11分)
∴-Sn+2Sn-1+2=0,即Sn=2Sn-1+2,
∴Sn+2=2(Sn-1+2). (13分)
∵S1+2=4≠0,∴Sn-1+2≠0,∴
Sn+2
Sn−1+2=2,(14分)
故{Sn+2}是以4为首项,2为公比的等比数列. (15分)
∴当n=1时,a1=2×1=2; (2分)
当n=2时,a1+2a2=(a1+a2)+4,∴a2=4; (5分)
当n=3时,a1+2a2+3a3=2(a1+a2+a3)+6,∴a3=8.(8分)
(2)证明:∵a1+2a2+3a3+…+nan=(n-1)Sn+2n(n∈N*),①
∴当n≥2时,a1+2a2+3a3+…+(n-1)an-1=(n-2)Sn-1+2(n-1).②(9分)
①-②得nan=(n-1)Sn-(n-2)Sn-1+2
∴nan=n(Sn-Sn-1)-Sn+2Sn-1+2
∴nan=nan-Sn+2Sn-1+2.(11分)
∴-Sn+2Sn-1+2=0,即Sn=2Sn-1+2,
∴Sn+2=2(Sn-1+2). (13分)
∵S1+2=4≠0,∴Sn-1+2≠0,∴
Sn+2
Sn−1+2=2,(14分)
故{Sn+2}是以4为首项,2为公比的等比数列. (15分)
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