Question

if a principal, P, is invested at r% interest compounded annually then its future value, S, after n years is given by
\(\large S=P(1+ \frac{r}{100})^n \)
Use this formula to show that if an interest rate of r% is compounded k times a year, then after t years
\(\large S=P(1+ \frac{r}{100k})^kt \)
b) show that if m=100k/r then the formula in part (1) can be written as S+P((1+1/m)^m)^rt/100

c) use the difinition
e=lim m>00(1+1/m)^m
to deduce that if the interest is compounded with every- increasing frequency ( that is conitnuously ) then s=Pe^rt/100

oo=infinity
thank you for hellping me

Answer 1

i need some help ^^

Answer 2

i am not sure what it means by "show" but if your interest rate is \(r\) then if it is compounded \(k\) times per year, each \(k\)th part of a year you get \(\frac{r}{k}\) interest.
then if the number of years is \(t\) there are \(kt\) compounding periods, giving you the formula you wrote above

Answer 3

i said let n= t and k=1 so anwser gona be s=p(1+r)^n

Answer 4

i am not sure too that why i am asking

Answer 5

the second part is a replacement
\(\large S=P(1+ \frac{r}{100k})^{kt} \)
put \(m=\frac{100k}{r}\) which makes \(\frac{1}{m}=\frac{r}{100k}\)

Answer 6

this 100 is annoying, \(r\) should be written as a number, not a percent. but if that is what is says, that is what you have to do

Answer 7

yea it like that written i know it is annoying

Answer 8

if you replace \(\frac{r}{100k}\) by \(\frac{1}{m}\) you get
\(\large S=P(1+ \frac{1}{m})^{kt} \)

Answer 9

so it gona be the final anwser S=p(1+1/m)^kt after replacing m=100k/r right?

Answer 10

that was wrong, sorry

Answer 11

kk how i suppose submit in my paper how i start to writte i understand what you say how i should writte in my paper^^

Answer 12

since \[m=\frac{100k}{r}\] you get
\[k=\frac{mr}{100}\]

Answer 13

to your exponent is
\[{kt}=\frac{mrt}{100}=m\times \frac{rt}{100}\]

Answer 14

this gives
\[S+P((1+1/m)^m)^{rt/100}\]

Answer 15

with an equal sign not a plus sign

Answer 16

then as the \(k\to \infty\) you have \(m\to \infty\) i.e. as the compounding period gets shorter and shorter, i.e. as the number of compounding periods gets larger and larger, you get
\[\lim_{m\to \infty}P(1+\frac{1}{m})^m)^{rt/100}\]\[=Pe^{rt/100}\]

Answer 17

i would write the reasoning in english and the algebra as algebra

Answer 18

that the anwser for c?

Answer 19

yes

Answer 20

well, b and c

Answer 21

b is algebra, c is taking the limit to get \(e\)

Answer 22

for b i write like that since
m=100kr
you get
k=mr100
kt=mrt100=m×rt100
S+P((1+1/m)m)rt/100

Answer 23

for c ) like that
then as the k→∞ you have m→∞ i.e. as the compounding period gets shorter and shorter, i.e. as the number of compounding periods gets larger and larger, you get
limm→∞P(1+1m)m)rt/100
=Pert/100

Answer 24

right?

Answer 25

i should righ like that satellite?