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, …
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
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 $\displaystyle n^{th}$ term.
. . $\displaystyle
43, 44, 50, 53, 65, 70, 88, 95, 119, \hdots$
Consider the two subsequences formed by odd $\displaystyle n$ and even $\displaystyle n.$
$\displaystyle \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: .$\displaystyle a_n \;=\;\frac{2n^2 - n +85}{2}$ .[1] .for odd $\displaystyle n.$
$\displaystyle \begin{array}{c|cccc}
\text{even }n & 2 & 4 & 6 & 8 \\ \hline
a_n & 44 & 53 & 70 & 95 \end{array}$
. . The generating function is: .$\displaystyle a_n \;=\;\frac{2n^2 - 3n + 86}{2}$ .[2] .for even $\displaystyle n.$
To make the two functions alternate:
. . (1) Multiply [1] by $\displaystyle \frac{1 - (-1)^n}{2}$
. . (2) Multiply [2] by $\displaystyle \frac{1 + (-1)^n}{2}$
. . (3) Add the two results.
We have: .$\displaystyle 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: .$\displaystyle \boxed{a_n \;=\;\frac{4n^2 - 4n + 171 + (-1)^n(1-2n)}{4}}$
The next two terms are:
. . $\displaystyle a_{10} \;=\;\frac{4(10^2) - 4(10) + 171 + (-1)^{10}(1-20)}{4} \;=\;\frac{400 - 40 + 171 - 19}{4} \;=$ . $\displaystyle \frac{512}{4} \;=\;\boxed{128}$
. . $\displaystyle a_{11} \;=\;\frac{4(11^2) - 4(11) + 171 + (-1)^{11}(1-22)}{4} \;=\;\frac{484 - 44 + 171 + 21}{4} \;=$ . $\displaystyle \frac{632}{4} \;=\;\boxed{158}$