simplification

**Rambo** · March 12th 2008, 09:46 AM

How can i count this easier???

2500*(1,0525)^5+2500*(1,0525)^4+2500*(1.0525)^3+25 00*(1.0525)^2+2500*(1,0525)^1= 14612.10382

How could i count this without writing allt this down?

**topsquark** · March 12th 2008, 10:31 AM

Originally Posted by Rambo

How can i count this easier???

2500*(1,0525)^5+2500*(1,0525)^4+2500*(1.0525)^3+25 00*(1.0525)^2+2500*(1,0525)^1= 14612.10382

How could i count this without writing allt this down?

The only way I can think of to help with this is to factor a bit:
$(2500 \cdot 10525)(10525^4 + 10525^3 + 10525^2 + 10525 + 1)$

-Dan

**Rambo** · March 12th 2008, 10:45 AM

Originally Posted by topsquark

The only way I can think of to help with this is to factor a bit:
$(2500 \cdot 10525)(10525^4 + 10525^3 + 10525^2 + 10525 + 1)$

-Dan

thats not correct that gets a other answer 17679.91359

**topsquark** · March 12th 2008, 10:51 AM

Originally Posted by Rambo

thats not correct that gets a other answer 17679.91359

Funny because I'm getting 322914659677371145062500 doing it both ways. Not what you got. How did you get such a small answer?

-Dan

Edit: Oh, I get it now. It isn't 10125 it is 1.0125. (The comma-decimal point thing gets me in the end again!) Still, I get the same answer using both expressions: 14612.103817994

-Dan

**mathceleb** · March 12th 2008, 10:52 AM

Rambo,

That is the Accumulated Value of an annuity due series. There is a standard formula for that:

Your Common Term or payment is 2500.

Your Sum of the Series:

Payment * ((1 + i)^n - 1) * (1 + i)/i

In your textbook, they might refer to d, which is i/(1 + i).

That being said, i is .0525, payment is 2500, and n is 5. Using that formula, I match your numbers to the decimal place.

**earboth** · March 12th 2008, 11:00 AM

Originally Posted by Rambo

How can i count this easier???

2500*(1,0525)^5+2500*(1,0525)^4+2500*(1.0525)^3+25 00*(1.0525)^2+2500*(1,0525)^1= 14612.10382

How could i count this without writing allt this down?

First I assume that you use the comma as a decimal point.

Second there must be a typo because the LHS yields 14612.10381

Third: you can factor out 2500:

$2500*(1.0525)^5+2500*(1.0525)^4+2500*(1.0525)^3+ 2500*(1.0525)^2+2500*(1.0525)^1 =$ $2500*(1.0525)^5+(1.0525)^4+(1.0525)^3+ (1.0525)^2+(1.0525)^1)= 14612.10381$

**Rambo** · March 12th 2008, 11:00 AM

2500*((1+0,525)^5*(1+0,525)/0,525= 59896.26856

The answer will be wrong!!!!

**mathceleb** · March 12th 2008, 11:06 AM

Originally Posted by Rambo

2500*((1+0,525)^5*(1+0,525)/0,525= 59896.26856

The answer will be wrong!!!!

Did you read it right? Payment * (1 + i) * ((1 + i)^n - 1)/i

2500 * 1.0525 * ((1.0525)^5 - 1)/.0525

This is 14612.10

It does what Earthbot's series does. You mentioned you like using series better, so I'd go with Earthbot's suggestion.

**wingless** · March 12th 2008, 11:33 AM

$2500(1,0525)^5+2500(1,0525)^4+2500(1.0525)^3+2500( 1.0525)^2+2500(1,0525)^1$

$2500 ({1,0525}^1+{1.0525}^2+{1.0525}^3+{1,0525}^4+{1,05 25}^5)$

$2500 \sum_{k=1}^{5} 1.0525^k$

$\sum_{k=1}^{n} r^k = \frac{r^{n+1}-1}{r-1}-1$

$2500 \sum_{k=1}^{5} 1.0525^k = 2500 \left (\frac{{1.0525}^{6}-1}{{1.0525}-1}-1 \right )$

$2500 \left ( \frac{0.35935}{0.0525}-1 \right ) = 14612.1$

**Soroban** · March 12th 2008, 12:46 PM

Hello, Rambo!

It helps if you know this identity:

. . $\frac{X^n - 1}{X - 1} \;=\;X^{n-1} + X^{n-2} + X^{n-3} + \hdots + X^2 + X+ 1$

Evaluate:
$2500(1.0525)^5+2500(1.0525)^4+2500(1.0525)^3+2500( 1.0525)^2+2500(1,0525)^1$

Factor: . $2500(1.0525)\,\bigg[1.0525^4 + 1.0525^3 + 1.0525^2 + 1.0525 + 1\bigg]$

. . . . $= \;2500(1.0525)\left[\frac{1.0525^5 - 1}{1.0525 - 1}\right]$

. . . . $= \;(2631.25)\,\frac{0.291547915}{0.0525}$

. . . . $= \;14612.10382$

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