求证1×2+2×3+3×4+…+n(n+1)=13n(n+1)(n+2)
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求证1×2+2×3+3×4+…+n(n+1)=
n(n+1)(n+2)
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![求证1×2+2×3+3×4+…+n(n+1)=13n(n+1)(n+2)](/uploads/image/z/13590282-66-2.jpg?t=%E6%B1%82%E8%AF%811%C3%972%2B2%C3%973%2B3%C3%974%2B%E2%80%A6%2Bn%28n%2B1%29%EF%BC%9D13n%28n%2B1%29%28n%2B2%29)
证明:①当n=1时,左边=2,右边=
1
3×1×2×3=2,等式成立;
②假设当n=k时,等式成立,
即1×2+2×3+3×4+…+k(k+1)=
1
3k(k+1)(k+2)
则当n=k+1时,
左边=
1
3k(k+1)(k+2)+(k+1)(k+2)=(k+1)(k+2)(
1
3k+1)=
1
3(k+1)(k+2)(k+3)
即n=k+1时,等式也成立.
所以1×2+2×3+3×4+…+n(n+1)=
1
3n(n+1)(n+2)对任意正整数都成立.
1
3×1×2×3=2,等式成立;
②假设当n=k时,等式成立,
即1×2+2×3+3×4+…+k(k+1)=
1
3k(k+1)(k+2)
则当n=k+1时,
左边=
1
3k(k+1)(k+2)+(k+1)(k+2)=(k+1)(k+2)(
1
3k+1)=
1
3(k+1)(k+2)(k+3)
即n=k+1时,等式也成立.
所以1×2+2×3+3×4+…+n(n+1)=
1
3n(n+1)(n+2)对任意正整数都成立.
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