作业帮已知数列{an}满足a1=2，an+1=an-1n(n+1)
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已知数列{an}满足a1=2，an+1=an-
| 1 |
| n(n+1) |
（Ⅰ）由an+1=an-
1
n(n+1)移向得an+1-an=-
1
n(n+1)=
1
n+1−
1
n
当n≥2时，an=（an-an-1）+（an-1-an-2）+…+（a2-a1）+a1
=（
1
n−
1
n−1）+（
1
n−1−
1
n−2）+…+（
1
2−
1
1）+2
=
1
n+1．
当n=1时，也适合上式，
所以数列{an}的通项公式为an=
1
n+1．
（Ⅱ）bn=nan•2n=（n+1）•2n，
sn=2×21+3×22+…+（n+1）×2n，①
2sn=2×22+3×23+…+n×2n+（n+1）×2n+1，②
两式相减得：
-sn=2×21+22+23…+2n-（n+1）×2n+1
=4+
22(1−2n−1)
1−2-（n+1）×2n+1
=2n+1-（n+1）×2n+1
=-n×2n+1．
则sn=n×2n+1．
1
n(n+1)移向得an+1-an=-
1
n(n+1)=
1
n+1−
1
n
当n≥2时，an=（an-an-1）+（an-1-an-2）+…+（a2-a1）+a1
=（
1
n−
1
n−1）+（
1
n−1−
1
n−2）+…+（
1
2−
1
1）+2
=
1
n+1．
当n=1时，也适合上式，
所以数列{an}的通项公式为an=
1
n+1．
（Ⅱ）bn=nan•2n=（n+1）•2n，
sn=2×21+3×22+…+（n+1）×2n，①
2sn=2×22+3×23+…+n×2n+（n+1）×2n+1，②
两式相减得：
-sn=2×21+22+23…+2n-（n+1）×2n+1
=4+
22(1−2n−1)
1−2-（n+1）×2n+1
=2n+1-（n+1）×2n+1
=-n×2n+1．
则sn=n×2n+1．
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