Pattern of Sequence

The first differences are $1,5,9,13$ .

The second differences are $4,4,4,4$ .

Since second differences are same, the function is quadratic.

Since $f(x)=x$ for $1,2$ .

$f(x) =k(x-1)(x-2)+x$

Putting value of $f(3)$ , $k=2$ .

Now, $f(100)=2(99)(98)+100 =\boxed{19504}$

We note that the first few differences between consecutive terms of the sequence are $1, 5, 9, 13, \cdots$ . Let the terms be $a_n$ , then for $n \ge 1$ we have:

$\begin{aligned} a_{n+1} - a_n & = 4(n-1) + 1 = 4n - 3 \\ \sum_{k=1}^n \left(a_{k+1} - a_k \right) & = 4\sum_{k=1}^n - \sum_{k=1}^n 3 \\ a_{n+1} - a_1 & = 2n(n+1) - 3n & \small \blue{\text{Replace }n+1 \text{ with }n.} \\ a_n - 1 & = 2(n-1)n - 3(n-1) \\ \implies a_n & = 2n^2 - 5n + 4 \\ \implies a_{100} & = 2(100^2) - 5(100) + 4 \\ & = \boxed{19504} \end{aligned}$

Here, difference between 2 and 1 = 1 Difference between 7 and 2 = 5 = 1+4×1 And between 16 and 7 = 9 = 1+4×2 And between 29 and 16 = 13 = 1+4×3 And so... However, I won't tell the ultimate formula. But, for the above sequence, the equation = 2n²–5n+4 Now, for 100th term = 2×100²–5×100+4 =19504