摘要:(3)由及①②两式易得∴的公差
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已知:n=
-
,n•(n+1)=
-
.
由以上两式,可以类比得到n(n+1)(n+2)=
-
-
.
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| n(n+1) |
| 2 |
| (n-1)•n |
| 2 |
| n•(n+1)•(n+2) |
| 3 |
| (n-1)•n•(n+1) |
| 3 |
由以上两式,可以类比得到n(n+1)(n+2)=
| n(n+1)(n+2)(n+3) |
| 4 |
| (n-1)•n•(n+1)(n+2) |
| 4 |
| n(n+1)(n+2)(n+3) |
| 4 |
| (n-1)•n•(n+1)(n+2) |
| 4 |