试证;对任意的正整数n,有1/(1*2*3)+1/(2*3*4)+.+1/n(n+1)(n+2)
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试证;对任意的正整数n,有1/(1*2*3)+1/(2*3*4)+.+1/n(n+1)(n+2)
![试证;对任意的正整数n,有1/(1*2*3)+1/(2*3*4)+.+1/n(n+1)(n+2)](/uploads/image/z/19380885-69-5.jpg?t=%E8%AF%95%E8%AF%81%EF%BC%9B%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E6%AD%A3%E6%95%B4%E6%95%B0n%2C%E6%9C%891%2F%281%2A2%2A3%29%2B1%2F%282%2A3%2A4%29%2B.%2B1%2Fn%28n%2B1%29%28n%2B2%29)
1/n(n+1)(n+2)=1/n·[1/(n+1) - 1/(n+2)]=1/n(n+1) - 1/n(n+2) =[1/n-1/(n+1)] - ½[1/n-1-(n+2)]
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]
=½[1/n-2/(n+1)+1/(n+2)].
∴原式=½(1/1-2/2+1/3)+½(1/2-2/3+1/4)+½(1/3-2/4+1/5)+···+½[1/n-2/(n+1)+1/(n+2)]
=½(1/1+1/2+1/3+···+1/n) - 1·[1/2+1/3+1/4+···+1/(n+1)] + ½[1/3+1/4+···+1/(n+2)]
=½[1+1/2+1/(n+1)+1/(n+2)] + 1·[1/3+1/4+···+1/n] - 1·[1/2+1/3+1/4+···+1/(n+1)]
=1/2+1/4 + ½[1/(n+1)+1/(n+2)] - [1/2+1/(n+1)]
=1/4-½[1/(n+1)-1/(n+2)]
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