设S1=1+112+122,S2=1+122+132,S3=1+132+142,…,Sn=1+1n2+1(n+1)2.
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设S
![设S1=1+112+122,S2=1+122+132,S3=1+132+142,…,Sn=1+1n2+1(n+1)2.](/uploads/image/z/15570878-14-8.jpg?t=%E8%AE%BES1%EF%BC%9D1%2B112%2B122%EF%BC%8CS2%EF%BC%9D1%2B122%2B132%EF%BC%8CS3%EF%BC%9D1%2B132%2B142%EF%BC%8C%E2%80%A6%EF%BC%8CSn%EF%BC%9D1%2B1n2%2B1%28n%2B1%292%EF%BC%8E)
∵Sn=1+
1
n2+
1
(n+1)2=
n2(n+1)2+(n+1)2+n2
n2(n+1)2=
[n(n+1)]2+2n2+2n+1
[n(n+1)]2=
[n(n+1)+1]2
[n(n+1)]2,
∴
Sn=
n(n+1)+1
n(n+1)=1+
1
n(n+1)=1+
1
n-
1
n+1,
∴S=1+1-
1
2+1+
1
2-
1
3+…+1+
1
n-
1
n+1
=n+1-
1
n+1
=
(n+1)2−1
n+1=
n2+2n
n+1.
故答案为:
n2+2n
n+1.
1
n2+
1
(n+1)2=
n2(n+1)2+(n+1)2+n2
n2(n+1)2=
[n(n+1)]2+2n2+2n+1
[n(n+1)]2=
[n(n+1)+1]2
[n(n+1)]2,
∴
Sn=
n(n+1)+1
n(n+1)=1+
1
n(n+1)=1+
1
n-
1
n+1,
∴S=1+1-
1
2+1+
1
2-
1
3+…+1+
1
n-
1
n+1
=n+1-
1
n+1
=
(n+1)2−1
n+1=
n2+2n
n+1.
故答案为:
n2+2n
n+1.
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