# A painting, purchased for $10 000 in 1990, increased in value by 8% per year. How do you find the value of the painting in the year 2000? ##### 1 Answer Jun 7, 2016 You have to compose the interests year by year. #### Explanation: Let's try to find the general law of this. The initial value of the paint, at zero years, is ${V}_{0}$(later we will substitute the$10,000).

After the first year we have a new value that is augmented of the percentage given. We call the percentage P (later we will substitute 8/100).

${V}_{1} = {V}_{0} + P \cdot {V}_{0} = {V}_{0} \left(P + 1\right)$.

The second year we have the same rule, but applied to ${V}_{1}$

${V}_{2} = {V}_{1} + P \cdot {V}_{1} = {V}_{1} \left(P + 1\right)$

and substituting the previous value of ${V}_{1} = {V}_{0} \left(P + 1\right)$ we have

${V}_{2} = {V}_{0} \left(P + 1\right) \left(P + 1\right) = {V}_{0} {\left(P + 1\right)}^{2}$.

It seems we found a rule, let's try again for the third year

${V}_{3} = {V}_{2} + P \cdot {V}_{2} = {V}_{2} \left(P + 1\right)$

and substituting the previous equation for ${V}_{2} = {V}_{0} {\left(P + 1\right)}^{2}$

${V}_{3} = {V}_{0} {\left(P + 1\right)}^{2} \left(P + 1\right) = {V}_{0} {\left(P + 1\right)}^{3}$.

Then, after $N$ years, we have

${V}_{N} = {V}_{0} {\left(P + 1\right)}^{N}$.

This is the rule that is governing the increases of the value in the years. Now it is enough to substitute

${V}_{0} = 10000$

$P = \frac{8}{100}$

$N = 2000 - 1990 = 10$

${V}_{N} = 10000 {\left(\frac{8}{100} + 1\right)}^{10}$

$= 10000 \cdot {\left(\frac{108}{100}\right)}^{10}$

$= 10000 \cdot {\left(1.08\right)}^{10}$

$\setminus \approx 10000 \cdot 2.16$

$= 21600$.

This paint was a good investment, after $10$ years it has a value that is more than double: \$21 600.