几道初三题1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4+…+n)主要是要过程,谢谢拉书上写的答案为2n/n+1
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![几道初三题1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4+…+n)主要是要过程,谢谢拉书上写的答案为2n/n+1](/uploads/image/z/11712696-24-6.jpg?t=%E5%87%A0%E9%81%93%E5%88%9D%E4%B8%89%E9%A2%981%2B1%2F%281%2B2%29%2B1%2F%281%2B2%2B3%29%2B1%2F%281%2B2%2B3%2B4%29%2B%E2%80%A6%E2%80%A6%2B1%2F%281%2B2%2B3%2B4%2B%E2%80%A6%2Bn%29%E4%B8%BB%E8%A6%81%E6%98%AF%E8%A6%81%E8%BF%87%E7%A8%8B%2C%E8%B0%A2%E8%B0%A2%E6%8B%89%E4%B9%A6%E4%B8%8A%E5%86%99%E7%9A%84%E7%AD%94%E6%A1%88%E4%B8%BA2n%2Fn%2B1)
几道初三题1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4+…+n)主要是要过程,谢谢拉书上写的答案为2n/n+1
几道初三题
1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4+…+n)主要是要过程,谢谢拉
书上写的答案为2n/n+1
几道初三题1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+……+1/(1+2+3+4+…+n)主要是要过程,谢谢拉书上写的答案为2n/n+1
和上一位回答的一样呀,把分母加起来就会发现:每一个分式都可以拆开为以二个相邻的数为分母,分子为一的分式的差,这样就可以一项一项抵消,最后剩下的加起来就是答案.注意,1/2化为(1-1/2)即可.
1+1/2+1/6+1/10..........+1/(n(n+1)/2)=1+(1-1/2)+(1/2-1/3)+.....+(1/n - 1/(n+1))
=2-1/(n+1)