分析:(Ⅰ)由a
n+1=a
n-
移向得出a
n+1-a
n=-
=
-,再利用疊加法求通項(xiàng).
(Ⅱ)(Ⅱ)b
n=na
n•2
n=(n+1)•2
n,根據(jù)通項(xiàng)公式特點(diǎn):等差數(shù)列和等比數(shù)列的乘積,利用錯(cuò)位相消法求和.
解答:解:(Ⅰ)由a
n+1=a
n-
移向得a
n+1-a
n=-
=
-當(dāng)n≥2時(shí),a
n=(a
n-a
n-1)+(a
n-1-a
n-2)+…+(a
2-a
1)+a
1=(
-)+(
-)+…+(
-)+2
=
+1.
當(dāng)n=1時(shí),也適合上式,
所以數(shù)列{a
n}的通項(xiàng)公式為a
n=
+1.
(Ⅱ)b
n=na
n•2
n=(n+1)•2
n,
s
n=2×2
1+3×2
2+…+(n+1)×2
n,①
2s
n=2×2
2+3×2
3+…+n×2
n+(n+1)×2
n+1,②
兩式相減得:
-s
n=2×2
1+2
2+2
3…+2
n-(n+1)×2
n+1=4+
-(n+1)×2
n+1=2
n+1-(n+1)×2
n+1=-n×2
n+1.
點(diǎn)評(píng):本題考查數(shù)列通項(xiàng)公式求解,數(shù)列求和,考查了裂項(xiàng)、疊加,錯(cuò)位相消法在數(shù)列問(wèn)題中的應(yīng)用體現(xiàn).