Question 1198943
Compound Interest Formula

{{{FV   =   P (1  +  r / n)^(Yn)}}}

where {{{P }}}is the starting principal, {{{r }}}is the annual interest rate, {{{Y }}}is the number of years invested, and {{{n}}} is the number of compounding periods per year. 
{{{FV}}} is the future value, meaning the amount the principal grows to after {{{Y}}} years.



a(annual)

{{{FV = P (1 + r/n)^(Y*n)}}}
{{{P=38800}}}
{{{r=0.05}}}
{{{n=1}}}
{{{Y=5}}}


{{{FV  = 38800 (1 + 0.05/1)^(5*1)}}}

{{{FV   =   38800 (1.0 )^5}}}

{{{FV   =   38800 (1.2762815625 )}}}

{{{FV   =   49519.72}}}



b(semiannual)

{{{FV   =   P (1  +  r / n)^(Y*n)}}}

{{{P=38800}}}
{{{r=0.05}}}
{{{n=2}}}
{{{Y=5}}}


{{{FV   =   38800 (1  + 0.05 / 2)^(5*2)}}}

{{{FV   =   38800 (1.025)^10}}}

{{{FV   =   38800 (1.2800845441963566)}}}

{{{FV   =   49667.28}}}



c(monthly)

{{{FV   =   P (1  +  r / n)^(Y*n)}}}

{{{P=38800}}}
{{{r=0.05}}}
{{{n=12}}}
{{{Y=5}}}


{{{FV   =   38800 (1  + 0.05 / 12)^(5*12)}}}

{{{FV   =   38800 (1.0041666666666667)^60}}}

{{{FV   =   38800 (1.2833586785)}}}

{{{FV   =   49794.32}}}



d(daily)

{{{FV   =   P (1  +  r / n)^(Y*n)}}}

{{{P=38800}}}
{{{r=0.05}}}
{{{n=365}}}
{{{Y=5}}}


{{{FV   =   38800 (1  + 0.05 / 365)^(5*365)}}}

{{{FV   =   38800 (1.0001369863013698)^1825}}}

{{{FV   =   38800 (1.28400343214692876)}}}

{{{FV   =   49819.33}}}