The next term problem is generally not a math problem, there is an infinity of possibilities, unless you spot a pattern. Possibly, starting from $k=0$, you have:
$$s_k = -(-2)^k.3^{1-k}\,,$$
or
$$s_k = -3 \left(\frac{-2}{3}\right)^k\,.$$
Works for the five first terms.
In your case, it is easy to spot powers of $2$ and powers of $3$, and the alternating sign. Powers suggest computing ratios of terms. The alternating sign suggests it is not too complicated, because $-1,1,-1,-1,\ldots$ is not.
Simple techniques are computing differences or ratios, or differences of differences, etc. And recognizing standard series. But this falls short fast with $5$ terms.
But you never know what the person who asks you the question has in mind. What would one do with $8,12,15,20,23,28,32,35,38,43,45,50,56$?
This is the cumulated number of letters from words of the lyrics of Shine on you crazy diamond, by the Pink Floyd, which I am currently listening to. So the next is $59$, because "now" follows "Remember when you were young, you shone like the sun. Shine on you crazy diamond".
Best Answer
$$\sum_{k=1}^{100}A(k){=\sum_{k=1}^{100}(k^2+1)k!\\=\sum_{k=1}^{100} (k^2+k-(k-1))k!\\= \sum_{k=1}^{100}k(k+1)!-(k-1)k!\\=100\times 101!}$$