Calculating the Net Present Value (NPV)
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When input the apropriate data is subbed into the equation we get:
$$NPV=sum_t=0^inftyfrac2001.1^t$$
I have been told that the second term looks like
$$sum_t=0^inftyfrac2001.1^t =frac2000.1=2000$$
Could someone explain how the second term can end up being written as $frac2000.1$?
sequences-and-series limits convergence finance geometric-series
edited Aug 27 at 12:58
Mefitico
65514
asked Aug 27 at 12:02
AniaN
63
add a comment |Â
up vote
1
down vote
favorite
When input the apropriate data is subbed into the equation we get:
$$NPV=sum_t=0^inftyfrac2001.1^t$$
I have been told that the second term looks like
$$sum_t=0^inftyfrac2001.1^t =frac2000.1=2000$$
Could someone explain how the second term can end up being written as $frac2000.1$?
sequences-and-series limits convergence finance geometric-series
edited Aug 27 at 12:58
Mefitico
65514
asked Aug 27 at 12:02
AniaN
63
What is the particular difficulty you're facing? You just plug in different values of $t$.
– Matti P.
Aug 27 at 12:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When input the apropriate data is subbed into the equation we get:
$$NPV=sum_t=0^inftyfrac2001.1^t$$
I have been told that the second term looks like
$$sum_t=0^inftyfrac2001.1^t =frac2000.1=2000$$
Could someone explain how the second term can end up being written as $frac2000.1$?
sequences-and-series limits convergence finance geometric-series
edited Aug 27 at 12:58
Mefitico
65514
asked Aug 27 at 12:02
AniaN
63
When input the apropriate data is subbed into the equation we get:
$$NPV=sum_t=0^inftyfrac2001.1^t$$
I have been told that the second term looks like
$$sum_t=0^inftyfrac2001.1^t =frac2000.1=2000$$
Could someone explain how the second term can end up being written as $frac2000.1$?
sequences-and-series limits convergence finance geometric-series
edited Aug 27 at 12:58
Mefitico
65514
asked Aug 27 at 12:02
AniaN
63
edited Aug 27 at 12:58
Mefitico
65514
edited Aug 27 at 12:58
Mefitico
65514
edited Aug 27 at 12:58
Mefitico
65514
65514
asked Aug 27 at 12:02
AniaN
63
asked Aug 27 at 12:02
AniaN
63
asked Aug 27 at 12:02
AniaN
63
63
What is the particular difficulty you're facing? You just plug in different values of $t$.
– Matti P.
Aug 27 at 12:07
add a comment |Â
What is the particular difficulty you're facing? You just plug in different values of $t$.
– Matti P.
Aug 27 at 12:07
What is the particular difficulty you're facing? You just plug in different values of $t$.
– Matti P.
Aug 27 at 12:07
What is the particular difficulty you're facing? You just plug in different values of $t$.
– Matti P.
Aug 27 at 12:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
Hint :
$frac 200 1.1^i$ is the general term of a geometric sequence.
answered Aug 27 at 12:05
nicomezi
3,6621819
add a comment |Â
up vote
2
down vote
When you sum a geometric series, you have
$$
S_N = sum_k=0^N a^k = 1 + a + a^2 + ldots + a^N\
aS_n = asum_k=0^N a^k = a + a^2 + a^3 + ldots + a^N+1
$$
now subtract both sides to get
$$
S_N - aS_N = 1 - a^N+1,
$$
which implies
$$
S_N = frac1-a^N+11-a.
$$
As $N to infty$, if $0 < a < 1$, we have $a^N+1 to 0$, so
$$
lim_N to infty S_N = frac11-a.
$$
Can you simplify your series to fit this result?
edited Aug 27 at 12:42
Botond
4,0082632
answered Aug 27 at 12:08
gt6989b
30.7k22248
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint :
$frac 200 1.1^i$ is the general term of a geometric sequence.
answered Aug 27 at 12:05
nicomezi
3,6621819
add a comment |Â
up vote
2
down vote
Hint :
$frac 200 1.1^i$ is the general term of a geometric sequence.
answered Aug 27 at 12:05
nicomezi
3,6621819
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint :
$frac 200 1.1^i$ is the general term of a geometric sequence.
answered Aug 27 at 12:05
nicomezi
3,6621819
Hint :
$frac 200 1.1^i$ is the general term of a geometric sequence.
answered Aug 27 at 12:05
nicomezi
3,6621819
answered Aug 27 at 12:05
nicomezi
3,6621819
answered Aug 27 at 12:05
nicomezi
3,6621819
answered Aug 27 at 12:05
nicomezi
3,6621819
3,6621819
add a comment |Â
add a comment |Â
up vote
2
down vote
When you sum a geometric series, you have
$$
S_N = sum_k=0^N a^k = 1 + a + a^2 + ldots + a^N\
aS_n = asum_k=0^N a^k = a + a^2 + a^3 + ldots + a^N+1
$$
now subtract both sides to get
$$
S_N - aS_N = 1 - a^N+1,
$$
which implies
$$
S_N = frac1-a^N+11-a.
$$
As $N to infty$, if $0 < a < 1$, we have $a^N+1 to 0$, so
$$
lim_N to infty S_N = frac11-a.
$$
Can you simplify your series to fit this result?
edited Aug 27 at 12:42
Botond
4,0082632
answered Aug 27 at 12:08
gt6989b
30.7k22248
add a comment |Â
up vote
2
down vote
When you sum a geometric series, you have
$$
S_N = sum_k=0^N a^k = 1 + a + a^2 + ldots + a^N\
aS_n = asum_k=0^N a^k = a + a^2 + a^3 + ldots + a^N+1
$$
now subtract both sides to get
$$
S_N - aS_N = 1 - a^N+1,
$$
which implies
$$
S_N = frac1-a^N+11-a.
$$
As $N to infty$, if $0 < a < 1$, we have $a^N+1 to 0$, so
$$
lim_N to infty S_N = frac11-a.
$$
Can you simplify your series to fit this result?
edited Aug 27 at 12:42
Botond
4,0082632
answered Aug 27 at 12:08
gt6989b
30.7k22248
add a comment |Â
up vote
2
down vote
up vote
2
down vote
When you sum a geometric series, you have
$$
S_N = sum_k=0^N a^k = 1 + a + a^2 + ldots + a^N\
aS_n = asum_k=0^N a^k = a + a^2 + a^3 + ldots + a^N+1
$$
now subtract both sides to get
$$
S_N - aS_N = 1 - a^N+1,
$$
which implies
$$
S_N = frac1-a^N+11-a.
$$
As $N to infty$, if $0 < a < 1$, we have $a^N+1 to 0$, so
$$
lim_N to infty S_N = frac11-a.
$$
Can you simplify your series to fit this result?
edited Aug 27 at 12:42
Botond
4,0082632
answered Aug 27 at 12:08
gt6989b
30.7k22248
When you sum a geometric series, you have
$$
S_N = sum_k=0^N a^k = 1 + a + a^2 + ldots + a^N\
aS_n = asum_k=0^N a^k = a + a^2 + a^3 + ldots + a^N+1
$$
now subtract both sides to get
$$
S_N - aS_N = 1 - a^N+1,
$$
which implies
$$
S_N = frac1-a^N+11-a.
$$
As $N to infty$, if $0 < a < 1$, we have $a^N+1 to 0$, so
$$
lim_N to infty S_N = frac11-a.
$$
Can you simplify your series to fit this result?
edited Aug 27 at 12:42
Botond
4,0082632
answered Aug 27 at 12:08
gt6989b
30.7k22248
edited Aug 27 at 12:42
Botond
4,0082632
edited Aug 27 at 12:42
Botond
4,0082632
edited Aug 27 at 12:42
Botond
4,0082632
4,0082632
answered Aug 27 at 12:08
gt6989b
30.7k22248
answered Aug 27 at 12:08
gt6989b
30.7k22248
answered Aug 27 at 12:08
gt6989b
30.7k22248
30.7k22248
add a comment |Â
add a comment |Â
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What is the particular difficulty you're facing? You just plug in different values of $t$.
– Matti P.
Aug 27 at 12:07