Question

Calculus II question!

Answer 1

We have no idea where to start, any help? PLease File is attached.

Answer 2

Start with subbing in the equation for Pn into Pn

Answer 3

There is only one equation I got this. \[\sum_{n=0}^{\infty} (\frac{ \lambda }{ \mu })^n(1-\frac{ \lambda }{ \mu })\]

Answer 4

\[\sum_{n=0}^{inf}r^n(1-r)=1\]
\[\sum_{n=0}^{inf}r^n-r^{n-1}=1\]
\[\sum_{n=0}^{inf}r^n-\sum_{n=0}^{inf}r^{n-1}=1\]
\[\sum_{n=0}^{inf}r^n=1+\sum_{n=0}^{inf}r^{n-1}\]
are some owrk abouts

Answer 5

since 0<|r|<1 the series converge and we can sub in some formulas

Answer 6

\[\sum_{n=0}^{inf}r^n-\sum_{n=0}^{inf}r^{n+1}=1\]
\[\frac{1}{1-r}-r\frac{1}{1-r}=1\]
\[\frac{1}{1-r}(1-r)=1\]

Answer 7

n-1 was a typo .... :)

Answer 8

since r is neither 1 nor 0 .... this is true for all r

Answer 9

Wait, all you did was distribute the r^n in the second step, should it be plus and not minus?

Answer 10

yeah, n+1 which i caught later ...

Answer 11

Oh okay!

Answer 12

And for the second part? how would we do that?

Answer 13

same concept ....

Answer 14

n (r^n(1-r) )

Answer 15

of we can see this as a telescoping sum that reduces to this

0
+r - r^2
+2r^2 - 2r^3
+3r^3 - 3r^4
+4r^4 - 4r^5
+5r^5 - 5 r^6
------------------------------------
r + r^2 + r^3 + r^4 + r^5 + .....

Answer 16

in other words: that formulas out to:\[\frac{r}{1-r}\]

Answer 17

do you understand where the formulas for an infinite geometric series come from?

Answer 18

Thanks so much man!