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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
3
[k(k+1)(k+2)-(k-1)k(k+1)],由此得1×2=
1
3
(1×2×3-0×1×2),2×3=
1
3
(2×3×4-1×2×3),…,n(n+1)=
1
3
[n(n+1)(n+2)-(n-1)n(n+1)].相加,得1×2+2×3+…+n(n+1)=
1
3
n(n+1)(n+2).
(1)类比上述方法,请你计算“1×2×3+2×3×4+…+n(n+1)(n+2)”的结果.
(2)试用数学归纳法证明你得到的等式.
人气:402 ℃ 时间:2019-08-21 09:34:31
解答
(1)∵n(n+1)(n+2)=14[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]∴1×2×3=14(1×2×3×4-0×1×2×3)2×3×4=14(2×3×4×5-1×2×3×4)…n(n+1)(n+2)=14[n(n+1)(n+2)(n+3)-(n-1)n(n+...
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