A plausible guess might be that the sequence is formed by a degree-4* polynomial,
[tex]x_n = a n^4 + b n^3 + c n^2 + d n + e[/tex]
From the given known values of the sequence, we have
[tex]\begin{cases}a+b+c+d+e = -2 \\ 16 a + 8 b + 4 c + 2 d + e = 1 \\ 81 a + 27 b + 9 c + 3 d + e = 7 \\ 256 a + 64 b + 16 c + 4 d + e = 25 \\ 625 a + 125 b + 25 c + 5 d + e = 79\end{cases}[/tex]
Solving the system yields coefficients
[tex]a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4[/tex]
so that the n-th term in the sequence might be
[tex]\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}[/tex]
Then the next few terms in the sequence could very well be
[tex]\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}[/tex]
It would be much easier to confirm this had the given sequence provided just one more term...
* Why degree-4? This rests on the assumption that the higher-order forward differences of [tex]\{x_n\}[/tex] eventually form a constant sequence. But we only have enough information to find one term in the sequence of 4th-order differences. Denote the k-th-order forward differences of [tex]\{x_n\}[/tex] by [tex]\Delta^{k}\{x_n\}[/tex]. Then
• 1st-order differences:
[tex]\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}[/tex]
• 2nd-order differences:
[tex]\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}[/tex]
• 3rd-order differences:
[tex]\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}[/tex]
• 4th-order differences:
[tex]\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}[/tex]
From here I made the assumption that [tex]\Delta^4\{x_n\}[/tex] is the constant sequence {15, 15, 15, …}. This implies [tex]\Delta^3\{x_n\}[/tex] forms an arithmetic/linear sequence, which implies [tex]\Delta^2\{x_n\}[/tex] forms a quadratic sequence, and so on up [tex]\{x_n\}[/tex] forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.
