Question

What is the value of the series:
sum of (3^n+1)/(4^n) for n=1 to infinity.I will also write the equation w/ the eqution box

Answer 1

teh equation box doesnt work

Answer 2

it keeps saying math procesing error

Answer 3

\[sum_{n=1}^{infty} (3^n+1)/ (4^n)\]

Answer 4

\[\sum_{n=1}^{\infty}\frac{3^n+1}{4^n}\] like that?

Answer 5

oh well
even if you refresh?

Answer 6

let me try

Answer 7

oh i see it now

Answer 8

no teh series is from n=1 to infinity
numerator - 3^ (n+1) .....the n+1 are together
denominator - 4^n

Answer 9

or maybe it is
\[\sum_{n=1}^{\infty}\frac{3^{n+1}}{4^n}\]

Answer 10

yes that correct

Answer 11

ok lets pull out a 3

Answer 12

\[3\sum_{n=1}^{\infty}\frac{3^{n}}{4^n}\]

Answer 13

what happedn to teh 1 that is with n+1

Answer 14

\[3\sum_{k=1}^{\infty}(\frac{3}{4})^n\]

Answer 15

i factored out a 3, precisely so the exponents would match

Answer 16

oh because 3^n * 3 is 3^n+1 right

Answer 17

and now it is easy enough right?

Answer 18

so now u use teh geometric series test right, with
\[\left| 3/4 \right| < 1 \]
so it converges

Answer 19

sure does

Answer 20

are you summing from 1 to infinity or from zero to infinity?

Answer 21

from 1 to infinity

Answer 22

ohh so if its from one teh exponent has to be n-1 though!

Answer 23

then the answer is
\[\frac{\frac{3}{4}}{1-\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3\]

Answer 24

oh and don't forget to multiply by the 3 out front, so your real answer is 9

Answer 25

wait why is a = 3/4

Answer 26

you are startinng at n = 1, not zero

Answer 27

so when u start from n=1, the a value is equal to the r value?

Answer 28

so you have
\[\frac{3}{4}+(\frac{3}{4})^2+(\frac{3}{4})^3+...\]
\[\frac{3}{4}(1+\frac{3}{4}+(\frac{3}{4})^2+..\]

Answer 29

so it looks like
\[a+ar+ar^2+ar^3+...\] where
\[a=\frac{3}{4}, r=\frac{3}{4}\] and you can use
\[\frac{a}{1-r}\]

Answer 30

so will it be safe to assume that when it is from n=1and its in the form ar^n, then the a and r value are the same

Answer 31

i guess. lets look at is a simpler way
if it was
\[\sum_{n=0}^{\infty}(\frac{3}{4})^n\] you would get
\[\frac{1}{1-\frac{3}{4}}=4\] but since you are starting at 1, you can subtract 1 from the result and get 3

Answer 32

omg that makes perfect sense now!! =D

Answer 33

yw

Answer 34

i did not even think of that

Answer 35

don't forget to multiply by 3 to get your real answer
good luck

Answer 36

i cant believe seris are soo easy yet sometimes i dont see teh things right before my eyes
and THANKS ALOT!!

Answer 37

yw (alot)

Answer 38

LOL

Answer 39

wait can u please help me with one more problem when i post it