已知數(shù)列{an}的前n項(xiàng)和Sn=n2.
(Ⅰ)求數(shù)列{an}的通項(xiàng)公式;
(Ⅱ)設(shè)an=2nbn,求數(shù)列{bn}的前n項(xiàng)和Tn.
分析:(Ⅰ)因?yàn)榻o出了數(shù)列{an}的前n項(xiàng)和Sn=n2,所以可用n≥2時(shí),an=sn-sn-1來(lái)求數(shù)列{an}的通項(xiàng)公式.
(Ⅱ)把(Ⅰ)中求出的數(shù)列{an}的通項(xiàng)公式代入an=2nbn,求出數(shù)列{bn}的通項(xiàng)公式,再利用錯(cuò)位相減法求數(shù)列{bn}的前n項(xiàng)和Tn.
解答:解:(Ⅰ)當(dāng)n=1時(shí),a
1=S
1=1,當(dāng)n>1時(shí),
an=Sn-Sn-1=n2-(n-1)2=2n-1,當(dāng)n=1時(shí)a
1=S
1=1
∴a
n=2n-1.
(Ⅱ)由
an=2nbn=2n-1,得bn=,
Tn=+++…+,①
2Tn=1+++…+,②
②-①,得
Tn=1+1+++…+-=3-.
點(diǎn)評(píng):本題主要考查數(shù)列通項(xiàng)公式與前n項(xiàng)和之間的關(guān)系,以及錯(cuò)位相減法求和.