Algebraic definition for the nth term

**warriors837** · September 15th 2009, 06:22 PM

Determine the next two numbers in the following sequence
and determine an explicit algebraic definition for the nth term.

43, 44, 50, 53, 65, 60, 88, 95, 119, …

**Calculus26** · September 15th 2009, 06:46 PM

are you sure its not
43,44,50,53,65,70,88,95,119 ?

the differences between terms follows the pattern

1,6,3,12,5,18,7,24 which means the differnce in the next 2 terms would be 9,30 giving 128 and 158 as the next 2 terms

the terms alternate increasing by 2n+1 and 6n

**warriors837** · September 16th 2009, 03:54 AM

No, it's 43, 44, 50, 53, 65, 60, 88, 95, 119, …

**warriors837** · September 16th 2009, 04:33 AM

But I sincerely feel that it was a mistake, because I found the same pattern, but was disgruntled when approaching the 60, so I think that it has to be 70.

**warriors837** · September 16th 2009, 04:40 AM

If this is so an it really is 70, then what would be the explicit algebraic definition for the nth term? And how do you do this?

**Soroban** · September 16th 2009, 10:12 AM

Hello, warriors8371

If the sixth term is 70, there is an interesting solution.

Determine the next two numbers in the following sequence
and determine an explicit algebraic definition for the $n^{th}$ term.

. . $ 43, 44, 50, 53, 65, 70, 88, 95, 119, \hdots$

Consider the two subsequences formed by odd $n$ and even $n.$

$\begin{array}{c|ccccc} \text{odd }n & 1 & 3 & 5 & 7 & 9 \\ \hline a_n & 43 & 50 & 65 & 88 & 119 \end{array}$

. . The generating function is: . $a_n \;=\;\frac{2n^2 - n +85}{2}$ .[1] .for odd $n.$

$\begin{array}{c|cccc} \text{even }n & 2 & 4 & 6 & 8 \\ \hline a_n & 44 & 53 & 70 & 95 \end{array}$

. . The generating function is: . $a_n \;=\;\frac{2n^2 - 3n + 86}{2}$ .[2] .for even $n.$

To make the two functions alternate:

. . (1) Multiply [1] by $\frac{1 - (-1)^n}{2}$

. . (2) Multiply [2] by $\frac{1 + (-1)^n}{2}$

. . (3) Add the two results.

We have: . $a_n \;=\;\frac{1-(-1)^n}{2}\left(\frac{2n^2-n+85}{2}\right) + \frac{1+(-1)^n}{2}\left(\frac{2n^2 - 3n + 86}{2}\right)$

. . which simplifies to: . $\boxed{a_n \;=\;\frac{4n^2 - 4n + 171 + (-1)^n(1-2n)}{4}}$

The next two terms are:

. . $a_{10} \;=\;\frac{4(10^2) - 4(10) + 171 + (-1)^{10}(1-20)}{4} \;=\;\frac{400 - 40 + 171 - 19}{4} \;=$ . $\frac{512}{4} \;=\;\boxed{128}$

. . $a_{11} \;=\;\frac{4(11^2) - 4(11) + 171 + (-1)^{11}(1-22)}{4} \;=\;\frac{484 - 44 + 171 + 21}{4} \;=$ . $\frac{632}{4} \;=\;\boxed{158}$

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