# Thread: Algebraic definition for the nth term

1. ## Algebraic definition for the nth term

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, …

2. 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

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

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.

5. 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?

6. 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:

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

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