Proving a formula: what is: 1 – X^(n+1) / (1 – X) when X = 1 / (1+i) ?

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__SB

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Proving a formula: what is: 1 – X^(n+1) / (1 – X) when X = 1 / (1+i) ?

Hopefully somebody can assist me with this.
Consider a sequence of equal payments, £z over n years(incl. this year).
Prove the value of these payments today can be written as
DPwB9g4Btd0z04PqgalgAAAABJRU5ErkJggg==

To begin;
The value today £V = Z + (Z * 1 / (1 + i)) + (Z * 1 / (1 +i)^2) +…
Next we note the geometric series: 1 + X + X^2 +…+ X^n can be written as
1 – X^(n+1) / (1 – X)
We can rewrite the value today as
£V = Z ( 1 + (1 / (1 + i)) + (1 / (1 + i)^2) +…)
We can therefore say that X in our geometric series = 1 / (1+ i)
The part I’m having trouble with is using the geometricresult;
i.e. what is: 1 – X^(n+1) / (1 – X) when X = 1 / (1+i)

Many thanks in advance.
 
** Can be written as: £V = Z * {1 – (1/(1+i)^n)} / {1 – (1/(1+i)}

 
__SB said:
Hopefully somebody can assist me with this.
Consider a sequence of equal payments, £z over n years(incl. this year).
Prove the value of these payments today can be written as

http://www.freemathhelp.com/forum/i...Nv/tmYAtmY6tpCQtv//25A625CQ29uQ2////7Zm/9uQ// 2///badx3nwAAAAF0Uk5TAEDm2GYAAAAJcEhZcwAADsQAAA7EAZUrDhsAAAAZdEVYdFNvZnR3YXJlAE1pY3Jvc29mdCBPZmZpY2V/7TVxAAACfUlEQVRYR 1YiVLDIBBNtIe2aqvGttoc// X7gGEMykCccaRcbQNu Tt24u1qgqstq7vz 65l9V1 mXN2thv6tp8EAu1f3r3qrSMbty wHuG48586FFlxW6zbmuyxQIcie/iZQkO4deM28PxAF8JnqmjG0hKLcI6fW3W/UsyvmYneHLMoufaNr6VoVg6jgPIkgrobsmzafwJnlzWBW4FvwHqiBnFrdAZ8Q3HGpGBAfjrAD8nkF8GX/eAdIpQNzjX BMfMYfwY0Nwc ED63ndUcJY/GF2MDE6f5ZO90j8SppZOBc y8UmvuGVI8/Ch181/siIX8HHlMz5F2IOl1kjU/ljdzpL8sfbzJzKD11n5K/fHz6IZqMiJ GDLiEDzudfuU3ZIV1p6Tj5y/VFrSR8Xuo4iPTuJgJL1Oewkif8mM1 j 0SLFtdIacCLlNyMy9w8CmKLGq85wiuDf6wL/bP56pBo7uNwa9 iJS7AaB P9AKRjN7PzAF H7QbTnIKXO4rfhtE3I34MsqIvFh0HCpmsZHIe67PmVFpQ5TGYX4 I7By 4JMluhUGE4LLYkfxCbosfM ZfhiY5W M8Yf4CPKuScfyeDoACrMsG67RtVG k7y7 qAUzFaAF4eNmqKmCu33OBCV36pdzC8Uf5gfkIHRDN11NEp0PKQVG6aWxJm5HCo9FkDHr8V2hGCl39IPw Y4pHqRkpiG9fazVwNt7LzUi5ri6TM5x2lYmekdJG87H/BFqdNYPEzkipo4eDzymXjFvSGz0jLevf BlpWXzxM1I2fIVmpDz4ys1IefCFC2TqjLQkvp/MSMXxaUNA/IyU/K/T2c73L/DPwB9g4Btd0z04PqgalgAAAABJRU5ErkJggg==
Click to expand...
Whatever your image was, it isn't displaying on my end. Sorry. :oops:

__SB said:
To begin;
The value today £V = Z + (Z * 1 / (1 + i)) + (Z * 1 / (1 +i)^2) +…
Next we note the geometric series: 1 + X + X^2 +…+ X^n can be written as
1 – X^(n+1) / (1 – X)
We can rewrite the value today as
£V = Z ( 1 + (1 / (1 + i)) + (1 / (1 + i)^2) +…)
We can therefore say that X in our geometric series = 1 / (1+ i)
The part I’m having trouble with is using the geometric result;
i.e. what is: 1 – X^(n+1) / (1 – X) when X = 1 / (1+i)
Click to expand...
It looks like all you're doing is plugging one formula in place of an expanded form. But that's a very simple process, so I must be misunderstanding your question. You say you have this:

. . . . .\(\displaystyle Z\, +\, \left(Z\, \cdot\, \dfrac{1}{(1\, +\, i)}\right)\, +\, \left(Z\, \cdot\, \dfrac{1}{(1\, +\, i)^2}\right)\, +\, ...\,+\, \left(Z\, \cdot\, \dfrac{1}{(1\, +\, i)^n}\right)\)

...which can be factored to be this:

. . . . .\(\displaystyle Z\,\bigg[1\, +\, \dfrac{1}{(1\, +\, i)}\, +\, \dfrac{1}{(1\, +\, i)^2}\, +\, ...\, +\, \dfrac{1}{(1\, +\, i)^n}\bigg]\)

You are given that:

. . . . .\(\displaystyle 1\, +\, x\, +\, x^2\, +\, ...\, +\, x^n\, =\, \dfrac{1\, -\, x^{n+1}}{1\, -\, x}\)

You say that you are using the following substitution:

. . . . .\(\displaystyle x\, =\, \dfrac{1}{1\, +\, i}\)

So your formula then becomes:

. . . . .\(\displaystyle Z\, \bigg[\dfrac{1\, -\, x^{n+1}}{1\, -\, x}\bigg]\, =\, Z\, \left[\dfrac{1\, -\, \dfrac{1}{(1\, +\, i)^{n+1}}}{1\, -\, \dfrac{1}{1\, +\, i}}\right]\, =\, Z\left[\dfrac{\left(\dfrac{(1\, +\, i)^{n+1}\, -\, 1}{(1\, +\, i)^{n+1}}\right)}{\left(\dfrac{(1\, +\, i)\, -\, 1}{1\, +\, i}\right)}\right]\, =\, ...\)

...and... then what? Where are you stuck? Please be specific. Thank you! ;)
 
Last edited:
stapel said:
Whatever your image was, it isn't displaying on my end. Sorry. :oops:


It looks like all you're doing is plugging one formula in place of an expanded form. But that's a very simple process, so I must be misunderstanding your question. You say you have this:

. . . . .\(\displaystyle Z\, +\, \left(Z\, \cdot\, \dfrac{1}{(1\, +\, i)}\right)\, +\, \left(Z\, \cdot\, \dfrac{1}{(1\, +\, i)^2}\right)\, +\, ...\,+\, \left(Z\, \cdot\, \dfrac{1}{(1\, +\, i)^n}\right)\)

...which can be factored to be this:

. . . . .\(\displaystyle Z\,\bigg[1\, +\, \dfrac{1}{(1\, +\, i)}\, +\, \dfrac{1}{(1\, +\, i)^2}\, +\, ...\, +\, \dfrac{1}{(1\, +\, i)^n}\bigg]\)

You are given that:

. . . . .\(\displaystyle 1\, +\, x\, +\, x^2\, +\, ...\, +\, x^n\, =\, \dfrac{1\, -\, x^n}{1\, -\, x}\)

You say that you are using the following substitution:

. . . . .\(\displaystyle x\, =\, \dfrac{1}{1\, +\, i}\)

So your formula then becomes:

. . . . .\(\displaystyle Z\, \bigg[\dfrac{1\, -\, x^n}{1\, -\, x}\bigg]\, =\, Z\, \left[\dfrac{1\, -\, \dfrac{1}{(1\, +\, i)^n}}{1\, -\, \dfrac{1}{1\, +\, i}}\right]\, =\, Z\left[\dfrac{\left(\dfrac{(1\, +\, i)^n\, -\, 1}{(1\, +\, i)^n}\right)}{\left(\dfrac{(1\, +\, i)\, -\, 1}{1\, +\, i}\right)}\right]\, =\, ...\)

...and... then what? Where are you stuck? Please be specific. Thank you! ;)
Click to expand...

Thanks for the reply.
I’m stuck in getting the final formula. You state that I amgiven 1 + x + X^2 +… = ( 1 – X^n) / (1 – x)
With that, I can get the final answer. However, my lecturersaid the series was equal to
(1 – X^(n+1)) / (1 –X)
With the n+1 power, I can’t seem to reach the equation.
 
__SB said:
I’m stuck in getting the final formula. You state that I amgiven 1 + x + X^2 +… = ( 1 – X^n) / (1 – x)
With that, I can get the final answer. However, my lecturersaid the series was equal to
(1 – X^(n+1)) / (1 –X)
With the n+1 power, I can’t seem to reach the equation.
Click to expand...
Apologies for the typo in the formula. Here's the correction:

stapel said:
So your formula then becomes:

. . . . .\(\displaystyle Z\, \bigg[\dfrac{1\, -\, x^{n+1}}{1\, -\, x}\bigg]\, =\, Z\, \left[\dfrac{1\, -\, \dfrac{1}{(1\, +\, i)^{n+1}}}{1\, -\, \dfrac{1}{1\, +\, i}}\right]\, =\, Z\left[\dfrac{\left(\dfrac{(1\, +\, i)^{n+1}\, -\, 1}{(1\, +\, i)^{n+1}}\right)}{\left(\dfrac{(1\, +\, i)\, -\, 1}{1\, +\, i}\right)}\right]\, =\, ...\)
Click to expand...
Please reply showing what you have done and where you're getting stuck. Thank you! ;)
 
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