http://keypress.com/documents/daa1/M...AA_MPYS_01.pdf
WORK SHEET 1.3 Problem #4 a-h
THANK YOU!
http://keypress.com/documents/daa1/M...AA_MPYS_01.pdf
WORK SHEET 1.3 Problem #4 a-h
THANK YOU!
Hello, 13373l1t3!
They asked for the "recursive" formula.
But if they want, say, the 21st term, we'd better have a "closed" form.
4. Write a recursive formula to generate each sequence.
Then find the indicated term.
$\displaystyle a)\;\;-15, -11, -7, -3, \hdots$ . Find the 10th term.
The sequence increases by 4 at each stage.
The recursive formula is: .$\displaystyle a_n \:=\:a_{n-1} + 4$
(Each term is the preceding term plus 4.)
The closed formula is: .$\displaystyle a_n \:=\:-19 + 4n$
Therefore: .$\displaystyle a_{10} \:=\:-19 + 4(10) \:=\:21$
$\displaystyle b)\;\;1000, 100, 10, 1, \hdots$ . Find the 12th term.
Each term is one-tenth of the preceding term.
The formula is: .$\displaystyle a_n \:=\:\frac{10,000}{10^n}$
Therefore: .$\displaystyle a_{12} \:=\:\frac{10,000}{10^{12} }\:=\:10^{-8} \:=\:0.00000001$
$\displaystyle c)\;\;17.25, 14.94, 12.63, 10.32, \hdots$ . Find the 15th term.
The terms decrease by 2.31
The formula is: .$\displaystyle a_n \:=\:19.56 - 2.31n$
Therefore: .$\displaystyle a_{15} \:=\:19.56 - 2.31(15) \:=\:-15.09$
$\displaystyle d)\;\;0.3, -0.03, 0.003, -0.0003,\hdots$ . Find the 8th term.
Each term is $\displaystyle -0.1$ times the preceding term.
The formula is: .$\displaystyle a_n \:=\:0.3(-0.1)^{n-1}$
Therefore: .$\displaystyle a_8 \;=\;0.3(-0.1)^7 \:=\:0.00000003$
$\displaystyle e)\;\;0, \frac{1}{6},\:\frac{1}{3},\:\frac{1}{2},\hdots$ . Find the 21st term.
This is a sneaky one!
Here is my thought process . . .
The numerators seems to be all 1's.
Then how come the first term has a numerator of 0 ?
Then it hit me . . . the numerators are: .$\displaystyle 0,1,2,3,4,\hdots$
. . and the denominators are all 6's !!
Get it? .We have: .$\displaystyle \frac{0}{6},\:\frac{1}{6},\:\frac{2}{6},\:\frac{3} {6},\:\frac{4}{6},\hdots $
. . and that is why the sequence begins: .$\displaystyle 0,\:\frac{1}{6},\:\frac{1}{3},\:\frac{1}{2},\:\hdo ts$
The formula is: .$\displaystyle a_n \;=\;\frac{n-1}{6}$
Therefore: .$\displaystyle a_{21} \;=\;\frac{21-1}{6} \:=\:\frac{20}{6} \:=\:\frac{10}{3}$
$\displaystyle f)\;\;-2, 4, -8, -16, \hdots$ . Find the 15th term.
Each term is $\displaystyle (-2)$ to a power.
The formula is: .$\displaystyle a_n \:=\:(-2)^n$
Therefore: .$\displaystyle a_{15} \;=\;(-2)^{15} \;=\;-32,768$