1/[a(a + 1)] + 1/[(a + 1)(a + 2)] + 1/[(a + 2)(a + 3)] + ………… + 1/[(a + 2004)(a + 2005)] 我们知道 1/a - 1/(a + 1) = [(a + 1) - a]/[a(a + 1)] = 1/[a(a + 1)],所以 原式= 1/a - 1/(a + 1) + 1/(a + 1) - 1/(a + 2) + 1/(a + 2) - 1/(a + 3) + ……… + 1/(a + 2004) - 1/(a + 2005) = 1/a - 1/(a + 2005) = 2005/[a(a + 2005)]