Question

Σ n = 1(start value) ∞(end value) 6 ( – 3 5 ) n(sequence)
a. – 9/4
b. 60/7
c. –9
d. 15 <---i choose this answer

Answer 1

-1 + x - x^2 + x^3 - x^4 + ....

this is a "tricky" series.


-1 + x
------------------------
1+x ) -1
(-1-x)
------
x
(x+x^2)
--------
-x^2

but this seems to be it:

-1/(1+x)

Answer 2

let x = 3/5

Answer 3

\[\sum_1^{inf}6(-\frac{3}{5})^n\]
\[6(3)\sum_1^{inf}(-\frac{1}{5})^n\]
\[6(3)~\frac 15(-\frac{1}{1+1/5})\]
\[6(3)~\frac 15(-\frac{1}{6/5})\]
\[\cancel 6(3)~\frac 1{\cancel 5}(-\frac{\cancel 5}{\cancel 6})=-3\]
gonna have to rethink this :)

Answer 4

the wolf says it aint d

Answer 5

ahh, i took out the 3 trying to be smart lol

Answer 6

:P

Answer 7

\[6~\frac 35(-\frac{1}{1+3/5})\]
\[-6~\frac 35(\frac{1}{8/5})\]
\[-6~\frac 35(\frac{5}{8})\]
\[-3~\frac 35(\frac{5}{4})\]
thats better

Answer 8

http://www.wolframalpha.com/input/?i=sum%28n%3D1..inf%29+%286%28-3%2F5%29%5En%29

now the wolf agrees :)

Answer 9

thanks!

Answer 10

how would i solve this question?
f(x) = 3x^2 – 5x; x_0 = 1
a. 4, 11, 19, 28, 38, 49

b. 4, 10, 16, 22, 28, 34

c. 4, 12, 21, 31, 42, 54

d. 4, 10, 18, 27, 37, 48

Answer 11

it seems to be just letting x=1,2,3,4,5 and finding the results

Answer 12

or, it seems that there might be some content of the question that is missing

Answer 13

oops sorry the correct multiple choices are:
a. –2, 7, 142
b.–1, –2, –11
c. 2, 5, 63
d. –2, –11, –38

Answer 14

well, when x=1, we get f=-2

but nothing else seems to fit.

Answer 15

f(x) = 3x^2 – 5x; x_0 = 1

this suggests that there is some iterative process going on but it is lacking any clear instructions to put it into context

Answer 16

its says to find the first three iterates of the function for the given initial value

Answer 17

well, -2,2,12 doesnt seem to be an option, and it doesnt seem clear yet that it is asking for a Newton method to find zeros ...

Answer 18

-2, 22 ,... would be the next reasonable approach as is, but even that is not an option

Answer 19

yeah i kept getting -2,22

Answer 20

f(x_0), f(f(x_0)), f(f(f(x_0))) is how an iterative process would work

f(1), f(f(1)), f(f(f(1)))

-2, f(-2), f(f(-2))

-2, 22, f(22)