Question

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

Answer 1

ok ready?

Answer 2

\[S_n=\sum_{k=1}^n kp^k=p+2p^2+3p^3+4p^4+...+np^n\]
\[pS_n=p^2+2p^3+3p^4+4p^5+...+np^{n+1}\]
\[S_n-pS_n=p+p^2+p^3+...+p^n-p^{n+1}\]
\[(1-p)S_n=\frac{1-p^{n+1}}{1-p}-p^{n+1}\]

Answer 3

divide by 1 - p get
\[S-n=\frac{1-p^{n+1}}{(1-p)^2}-\frac{p^{n+1}}{1-p}\] now take the limit as n goes to infinity

Answer 4

dammit typo, sum should be
\[(1-p)S_n=\frac{p-p^{n+1}}{1-p}-p^{n+1}\]