5.已知數(shù)列{an}滿足a1=1,an+1-an=2n(n∈N*),
(1)求數(shù)列{an}的通項(xiàng)公式;
(2)設(shè)bn=an+1,求數(shù)列{bn}的前n項(xiàng)和Tn.
分析 (1)由數(shù)列遞推式直接利用累加法求得數(shù)列{an}的通項(xiàng)公式;
(2)把數(shù)列{an}的通項(xiàng)公式代入bn=an+1,然后利用等比數(shù)列的前n項(xiàng)和求得數(shù)列{bn}的前n項(xiàng)和Tn.
解答 解:(1)由an+1-an=2n ,且a1=1,
得a2−a1=21,a3−a2=22,a4−a3=23,…,an−an−1=2n−1(n≥2),
累加得:an=a1+21+22+…+2n−1=1+2+…+2n−1=1×(1−2n)1−2=2n−1(n≥2),
當(dāng)n=1時(shí),上式成立,
∴an=2n−1;
(2)bn=an+1=2n-1+1=2n,
∴Tn =21+22+…+2n=2×(1−2n)1−2=2n+1−2.
點(diǎn)評(píng) 本題考查數(shù)列的求和,訓(xùn)練了累加法求數(shù)列的通項(xiàng)公式,考查了等比數(shù)列前n項(xiàng)和的求法,是中檔題.