若an=log(n+1)(n+2)(n∈N),我们把使乘积a1*a2……an为整数的数n叫做傲数,在区间（1,2011）内所有傲数和为
(n + 1)是对数的底吧?如果是,可以这么考虑.
先换底,an=log(n+2)/log(n+1)(n∈N),这里的log是以2为底的
于是,a1*a2……an
= log3/log2 * log4/log3 * ...* log(n+2)/log(n+1)
= log(n+2)/log2
= log(n+2)
也就是说,只有n = 2^k - 2时,n是傲数
1 < 2^k - 2 < 2011
1 < k < 11,即k = 2,3,4,5,6,7,8,9,10.共9个
这些傲数的和为
2^2 + 2^3 + ...+ 2^10 - 2*9
= 2^2 + (2^2 + 2^3 + ...+ 2^10) - 18 - 4
= 2^11 - 22
= 2048 - 22
= 2026
只是为了说明方法,不保证计算结果的正确性