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在计算“1×2+2×3+…+n(n+1)”时,先改写第k项:k(k+1)=13[k(k+1)(k+2)-(k-1)k(k

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在计算“1×2+2×3+…+n(n+1)”时,先改写第k项:k(k+1)=
1
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在计算“1×2+2×3+…+n(n+1)”时,先改写第k项:k(k+1)=13[k(k+1)(k+2)-(k-1)k(k
(1)∵n(n+1)(n+2)=
1
4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
∴1×2×3=
1
4(1×2×3×4-0×1×2×3)
2×3×4=
1
4(2×3×4×5-1×2×3×4)

n(n+1)(n+2)=
1
4[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]
∴1×2×3+2×3×4+…+n(n+1)(n+2)=
1
4[(1×2×3×4-0×1×2×3)+(2×3×4×5-1×2×3×4)+…+n×(n+1)×(n+2)×(n+3)-(n-1)×n×(n+1)×(n+2)=
1
4n(n+1)(n+2)(n+3)
(2)利用数学归纳法证:1×2×3+2×3×4+…+n(n+1)(n+2)=
1
4n(n+1)(n+2)(n+3)
①当n=1时,左边=1×2×3,右边=
1
4×1×2×3×4=1×2×3,左边=右边,等式成立.
②设当n=k(k∈N*)时,等式成立,
即1×2×3+2×3×4+…+k×(k+1)×(k+2)=
k(k+1)(k+2)(k+3)
4.  
则当n=k+1时,
左边=1×2×3+2×3×4+…+k×(k+1)×(k+2)+(k+1)(k+2)(k+3)
=
k(k+1)(k+2)(k+3)
4+(k+1)(k+2)(k+3)
=(k+1)(k+2)(k+3)(
k
4+1)
=
(k+1)(k+2)(k+3)(K+4)
4
=
(k+1)(k+1+1)(k+1+2)(k+1+3)
4.
∴n=k+1时,等式成立.
由①、②可知,原等式对于任意n∈N*成立.