用数学归纳法证明:1×2×3+2×3×4+…+n×(n+1)×(n+2)=n(n+1)(n+2)(n+3)4(n∈N
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用数学归纳法证明:1×2×3+2×3×4+…+n×(n+1)×(n+2)=
(n∈N
n(n+1)(n+2)(n+3) |
4 |
证明:(1)当n=1时,左边=1×2×3=6,右边=
1×2×3×4
4=6=左边,∴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.
1×2×3×4
4=6=左边,∴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.
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