求值1+1/2+2/2+1/2+1/3+2/3+3/3+2/3+1/3+...+1/100+2/100+...+2/100+1/100
问题描述:
求值1+1/2+2/2+1/2+1/3+2/3+3/3+2/3+1/3+...+1/100+2/100+...+2/100+1/100
答
因为 an = 1/n + 2/n +3/n + .+ (n-1)/n + n/n + (n-1)/n +.+1/n = 2[(1+2+3+...+(n-1)+n)] / n - 1 -----------先把中间数字算两次,然后扣除1次= 2 [n(n+1)/2] / n -1= (n+1) -1=n原式 = 1 + 1/2+2/2+1/2 + 1/3+2/3...