Series and Sequences Problem Part (a)

$\begin{aligned} f(1) & = 2 & ...(1) \\ f(2) - f(1) & = 3 & ...(2) \\ f(3) - f(2) & = 4 & ... (3) \\ \cdots & = \cdots & \cdots \\ f(n) - f(n-1) & = n+1 & ...(n) \end{aligned}$

Adding all the $n$ equations together $(1)+(2)+(3)+\cdots+(n)$ :

$\begin{aligned} \implies f(n) & = \color{#3D99F6} 2 + 3 + 4 + \cdots + n + (n+1) & \small \color{#3D99F6} \text{Sum of arithmetic progression} \\ & = {\color{#3D99F6}\frac {n(2+n+1)}2} = \frac {n(n+3)}2 & \small \color{#3D99F6} \implies S_n = \frac {n(a+l)}2 \\ \implies f(1000) & = \frac {1000(1003)}2 = \boxed{501500} \end{aligned}$

Well, in the series; we know the solution is going to be a quadratic. So using (n^2)/2 + an + b and f(1) = 2 as well as f(2) = 5. We can get a simultaneous equation for a and b. From there we find a = 3/2 and b = 0. Thus the general solution is n^2/2 + 3n/2.

Alternatively. If you know the triangular formula : n(n+1)/2 and add n to this; you get the same solution. This may take some trial and error, but can be derived logically by analyzing the pattern and noting that the first term is 2.

When you evaluate f(1000), in n^2/3 + 3n/2 you get the solution of 501500