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Gwar [14]
2 years ago
5

Determine the formula for the nth term of the sequence:-2,1,7,25,79,...​

Mathematics
1 answer:
rodikova [14]2 years ago
5 0

A plausible guess might be that the sequence is formed by a degree-4* polynomial,

x_n = a n^4 + b n^3 + c n^2 + d n + e

From the given known values of the sequence, we have

\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}

Solving the system yields coefficients

a=\dfrac58, b=-\dfrac{19}4, c=\dfrac{115}8, d = -\dfrac{65}4, e=4

so that the n-th term in the sequence might be

\displaystyle x_n = \boxed{\frac{5 n^4}{8}-\frac{19 n^3}{4}+\frac{115 n^2}{8}-\frac{65 n}{4}+4}

Then the next few terms in the sequence could very well be

\{-2, 1, 7, 25, 79, 208, 466, 922, 1660, 2779, \ldots\}

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 \{x_n\} 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 \{x_n\} by \Delta^{k}\{x_n\}. Then

• 1st-order differences:

\Delta\{x_n\} = \{1-(-2), 7-1, 25-7, 79-25,\ldots\} = \{3,6,18,54,\ldots\}

• 2nd-order differences:

\Delta^2\{x_n\} = \{6-3,18-6,54-18,\ldots\} = \{3,12,36,\ldots\}

• 3rd-order differences:

\Delta^3\{x_n\} = \{12-3, 36-12,\ldots\} = \{9,24,\ldots\}

• 4th-order differences:

\Delta^4\{x_n\} = \{24-9,\ldots\} = \{15,\ldots\}

From here I made the assumption that \Delta^4\{x_n\} is the constant sequence {15, 15, 15, …}. This implies \Delta^3\{x_n\} forms an arithmetic/linear sequence, which implies \Delta^2\{x_n\} forms a quadratic sequence, and so on up \{x_n\} forming a quartic sequence. Then we can use the method of undetermined coefficients to find it.

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