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?
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
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.
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:
$\displaystyle 2500*(1.0525)^5+2500*(1.0525)^4+2500*(1.0525)^3+ 2500*(1.0525)^2+2500*(1.0525)^1 = $ $\displaystyle 2500*(1.0525)^5+(1.0525)^4+(1.0525)^3+ (1.0525)^2+(1.0525)^1)= 14612.10381$
$\displaystyle 2500(1,0525)^5+2500(1,0525)^4+2500(1.0525)^3+2500( 1.0525)^2+2500(1,0525)^1$
$\displaystyle 2500 ({1,0525}^1+{1.0525}^2+{1.0525}^3+{1,0525}^4+{1,05 25}^5)$
$\displaystyle 2500 \sum_{k=1}^{5} 1.0525^k$
$\displaystyle \sum_{k=1}^{n} r^k = \frac{r^{n+1}-1}{r-1}-1$
$\displaystyle 2500 \sum_{k=1}^{5} 1.0525^k = 2500 \left (\frac{{1.0525}^{6}-1}{{1.0525}-1}-1 \right )$
$\displaystyle 2500 \left ( \frac{0.35935}{0.0525}-1 \right ) = 14612.1$
Hello, Rambo!
It helps if you know this identity:
. . $\displaystyle \frac{X^n - 1}{X - 1} \;=\;X^{n-1} + X^{n-2} + X^{n-3} + \hdots + X^2 + X+ 1$
Evaluate:
$\displaystyle 2500(1.0525)^5+2500(1.0525)^4+2500(1.0525)^3+2500( 1.0525)^2+2500(1,0525)^1$
Factor: .$\displaystyle 2500(1.0525)\,\bigg[1.0525^4 + 1.0525^3 + 1.0525^2 + 1.0525 + 1\bigg] $
. . . . $\displaystyle = \;2500(1.0525)\left[\frac{1.0525^5 - 1}{1.0525 - 1}\right]$
. . . . $\displaystyle = \;(2631.25)\,\frac{0.291547915}{0.0525}$
. . . . $\displaystyle = \;14612.10382$