Question

奥数题 ！！！！！！！！！！

假定:1Δ2=1/2+1/6, 1/2Δ3=1/6+1/12+1/20, 1/4Δ=1/20+1/30+1/42+1/56, 求1/2Δ2011的值

答案是2011/4026 求过程
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Answer 1

解:

1Δ2=1/2+1/6 = 1/(1×2) + 1/(2×3) = (1 - 1/2) + (1/2 - 1/3) = 1 - 1/3
同理
1/2Δ3=1/6+1/12+1/20 = 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 = 1/2 - 1/5
1/4Δ=1/20+1/30+1/42+1/56 = 1/4 - 1/8
所以
1/2Δ2011 = 1/2 - 1/2013 = (2013 - 2) / 4026 = 2011/4026

Answer 2

由题可得通项
1/nΔ(n+1)=1/n(n+1)+1/(n+1)(n+2)+……+1/(2n)(2n+1)
=1/n-1/(2n+1)

1/2001Δ2002=1/2001-1/4003