Question

calculus question help! picture included. for (b) it should be S=p/(1-p)^2

Answer 1

\[S=\frac{p}{(1-p)^2}\] yes?

Answer 2

yes!

Answer 3

snappier way to do this is to note that the derivative of
\[\frac{1}{1-x}\] is
\[\frac{1}{(1-x)^2}\] and therefore since
\[\sum x^n=\frac{1}{1-x}, |x|<1\] you get
\[\sum nx^{n-1}=\frac{1}{(1-x)^2}\] making
\[\sum nx^n=\frac{x}{(1-x)^2}\]

Answer 4

but that is not what you are asked to do, so we can do what you are asked

Answer 5

\[\S_{n} -\S_{n}p = p-p ^{n+1}\]
so
\[\S _{n}(1-p)= p-p ^{n+1}\]
\[\S _{n}=p-p ^{n+1}/1-p\]
now simplifie and take the limits

Answer 6

first we can use the integral test to show that it converges

Answer 7

actually this integral test is not so snappy, but i think we can use it, did you try?

Answer 8

and there is a small mistake in what myko wrote above

Answer 9

i dont understand can you write all the steps out?

Answer 10

sure
are you still here?

Answer 11

yes i really need the steps for a to c :( its hard for me you can shorten the steps tho

Answer 12

ok lets first two part b because you are given the instructions on what to do

Answer 13

\[S_n=\sum_{k=1}^nkp^k=p+2p^2+3p^3+...+np^n\]

Answer 14

\[pS_n=p^2+2p^3+3p^4+4p^5+...+np^{n+1}\]

Answer 15

subtract and get
\[S_n-pS_n=p+p^2+p^3+p^4+...+p^n-np^{n+1}\]

Answer 16

now
\[p+p^2+p^3+...+p^n=p(1+p+p^2+p^3+...+p^{n-1})=\frac{p-p^{n}}{1-p}\] by the formula for summing a geometric series

Answer 17

i migth be off by an exponent, hold on and let me check

Answer 18

yes, i am but it doesn't really matter, should be
\[p+p^2+p^3+...+p^n=\frac{p-p^{n+1}}{1-p}\]

Answer 19

subtract on the right, get
\[(1-p)S_n=\frac{p-(n-1)p^{n+1}+np^{n+2}}{1-p}\]
\[S_n=\frac{p-(n-1)p^{n+1}+np^{n+2}}{(1-p)^2}\] now take the limit as n goes to infinity, and since
\[0<p<1\] you get
\[S=\frac{p}{(1-p)^2}\]as desired

Answer 20

i have to say this is an amazingly cumbersome way to do a not very difficult problem

Answer 21

now that we have the formula part c is easy
replace p by 1/2 and compute
\[\frac{\frac{1}{2}}{(1-\frac{1}{2})^2}=\frac{\frac{1}{2}}{\frac{1}{4}}=2\]

Answer 22

what is the answer for a?:)