Question

Can someone please help me for a second? If so could you give this question a try:
Find the sum of the infinite geometric series, if it exists. 4 - 1 + 1/4 - + 1/16 . . .

A. - 1

B. 3

C. 16/5

D. does not exist

Answer 1

You multiply by -1/4 each time so that's the common ratio, r. The first term (a) is 4. The sum of an infinite geometric series is given by the following equation.\[4-1+\frac{1}{4}-\frac{1}{16}+ ... =\frac{a}{1-r}=\frac{4}{1-(-1/4)}=\frac{4}{5/4}=\frac{16}{5} \]So C

Answer 2

thank you soooooo much!

Answer 3

So, a(1) = 4 and subsequent terms are a(n+1) = (-1/4) a(n)?

a(1)/(1-x) = a1(1-x+x^2-x^3...) for x^2 < 1, so if x = (1/4)

your sum is 4/(1-1/4) = 4/(3/4) = 3

Got this from Standard Math Tables. You should check, as I could be wrong.

Answer 4

Evaluate the expression.

1 - 6p4/8p6

A. 1/56
B. 57/56
C. 13/28
D. 55/56

Answer 5

Can someone please give this one a try too?

Answer 6

@douglaswinslowcooper r is -1/4, not 1/4 so when you subtract it, it's positve.

Answer 7

Not sure what that is, do you mean this?\[1-\frac{6^4}{8^6}\]

Answer 8

I chose the form for 1-r so that I could use the sum with the alternating signs and r=1/4. Are you sure you are right?