Question

function f, f(1) = 2, f(x+1) = f(x)-1, f(100)=?

Answer 1

isnt this just recurrsion?

Answer 2

\[a_n=a_{n-1}-1\]

Answer 3

2 - 1(0)
2 - 1(1)
2 - 1(2)
2 - 1(3)
....
2 - 1(100)

Answer 4

may 2-1(99)

Answer 5

I don't get how it works

Answer 6

its like working in an office, youjust do the same thing over and over and over again

Answer 7

f(1) = 1
f(x+1) = f(x) - 1

f(1+1) = f(1) - 1; f(2) = 2-1 = 1

f(2+1) = f(2) - 1 = 1 - 1 = 0


etc ....

Answer 8

well, f(1) = 2 .... wrong key lol

Answer 9

f(1)=2, not 1

Answer 10

you can either do that 100 times, or develop the pattern:

f(x) = 2 - (x-1)

Answer 11

why f(x)=2-(x-1)? sorry..

Answer 12

its a pattern that develops over the course of the repetitions ..

f(1) = 2
f(2) = f(1) - 1= 2 - 1
f(3) = f(2) - 1= f(1) -1 - 1 = 2 - 1 - 1
f(4) = f(3) - 1= f(1) -1 - 1 - 1 = 2 - 1 - 1 - 1
....

f(n) = 2 - (1 - 1 - 1 - ... - 1), n-1 times
f(n) = 2 - (n-1)

Answer 13

right, I get it, thanks!

Answer 14

youre welcome

Answer 15

can you help with other AP question?

Answer 16

maybe one more

Answer 17

guy A walks 10km/day, the guy B walks 8km the first day and adds 0.5 each day to reach the guy A, how many days it took?

Answer 18

i assume B is trying to catch upto A?

Answer 19

yes

Answer 20

lets assume the start at the same place, when does their distances equal out?

d1: 10: 8
d2: 20: 8 + 8 + .5
d3: 30: 8 + 8 + .5 + 8 + .5 + .5
d4: 40: 8 + 8 + .5 + 8 + .5 + .5 + 8 + .5 + .5 + .5

or written another way

d1: 10: 8
d2: 2(10): 2(8) + .5
d3: 3(10): 3(8) + .5 + .5 + .5
d4: 4(10): 4(8) + .5 + .5 + .5 + .5 + .5 + .5

dn: n(10) = n(8) + 1(.5) + 2(.5) + 3(.5)+ ... + (n-1)(.5)


2n = .5(1+2+3+...+(n-1))

4n = 1+2+3+...+(n-1)
\[4n = \frac{(n-1)(n)}{2}\]\[4 = \frac{n-1}{2}\]\[8 = n-1\]\[9 = n\]
so my guess is in 9 days

Answer 21

after 8 days, that 80 to 79, so on the 9th day they catch up

Answer 22

where did that 2n = .5(1+2+3+...+(n-1)) come from?

Answer 23

0.5 is the common factor..

Answer 24

from the information given, each day the speed is .5 faster than the day before

day : distance covered
1 8
2 8 + .5
3 8 + .5 + .5
4 8 + .5 + .5 + .5

the distance covered over 4 days is the sum of all the days

4(8) + 0(.5) + 1(.5) + 2(.5) + 3(.4)

4(8) + .5 (1 + 2 + 3)

on the nth day

8n + .5 (1 + 2 + 3 + ... + n-1)

Answer 25

since the distance covered by each guy has to be equal for them to be at the same place; the other guy is walking 10 each day; 10n

10n = 8n + .5(....)

2n = .5(....)

4n = (....)

Answer 26

I see

Answer 27

then it's just algebra, ok, thanks again!

Answer 28

its always just algebra ;) good luck

Answer 29

=]