Suppose $$a_1, a_2, a_3, a_4, \ldots, a_n$$ is a sequence of numbers with the property that $$a_1 + a_2 + a_3 + \cdots + a_n = \frac{n(n+1)(n+2)}{6}$$, for all natural numbers $$n$$. [1] Find $$a_{100}$$, [2] Evaluate $$\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \cdots + \frac{1}{a_{89}} + \frac{1}{a_{90}}$$.

For $$n \ge 2$$, $$a_n = \frac{n(n+1)(n+2)}{6} - \frac{(n-1)n(n+1)}{6} = \frac{n(n+1)}{2}$$, and this also gives $$a_1 = 1$$; so the terms are $$1, 3, 6, 10, 15, \ldots$$. [1] $$a_{100} = \frac{100 \times 101}{2} = 5050$$. [2] Since $$\frac{1}{a_n} = \frac{2}{n(n+1)} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$$, the sum telescopes to $$2\left(1 - \frac{1}{91}\right) = \frac{180}{91}$$. So $$a_{100} = 5050$$ and the required sum is $$\frac{180}{91}$$.