#1**+10 **

a). Substitute 1 and 2 into the expression for R(n) to get two simultaneous equations for h and k:

n = 1: h^{1/2}/k = 4.8 (working on millions of dollars)

n = 2: h^{1}/k = 5.76 from this we have h = 5.76k. Put this into the first expression above to get:

(5.76k)^{1/2}/k = 4.8 or 5.76^{1/2}/k^{1/2} = 4.8. Multiply both sides by k^{1/2} and divide both sides by 4.8 to get:

k^{1/2} = 5.76^{1/2}/4.8. Square both sides: k = 5.76/4.8^{2}

$${\mathtt{k}} = {\frac{{\mathtt{5.76}}}{{{\mathtt{4.8}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.25}}$$

Hence from the n = 2: equation above we have h = 5.76*0.25 or

$${\mathtt{h}} = {\mathtt{5.76}}{\mathtt{\,\times\,}}{\mathtt{0.25}} \Rightarrow {\mathtt{h}} = {\mathtt{1.44}}$$

*Remember that the result of using these values in R(n) = h ^{n/2}/k will be in millions of dollars.*

For the sum of R(n) to n, notice that it can be written as:

sum(n) = (a + a^{2} + a^{3} + ... + a^{n})/k where a = h^{1/2}

The numerator is a geometric progression giving the result of sum(n) = a*(a^{n}-1)/[(a-1)k]

or sum(n) = (h^{1/2})*(h^{n/2}-1)/[(h^{1/2}-1)k}

See if you can now have a go at part b).

.

Alan Feb 23, 2015

#1**+10 **

Best Answer

a). Substitute 1 and 2 into the expression for R(n) to get two simultaneous equations for h and k:

n = 1: h^{1/2}/k = 4.8 (working on millions of dollars)

n = 2: h^{1}/k = 5.76 from this we have h = 5.76k. Put this into the first expression above to get:

(5.76k)^{1/2}/k = 4.8 or 5.76^{1/2}/k^{1/2} = 4.8. Multiply both sides by k^{1/2} and divide both sides by 4.8 to get:

k^{1/2} = 5.76^{1/2}/4.8. Square both sides: k = 5.76/4.8^{2}

$${\mathtt{k}} = {\frac{{\mathtt{5.76}}}{{{\mathtt{4.8}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.25}}$$

Hence from the n = 2: equation above we have h = 5.76*0.25 or

$${\mathtt{h}} = {\mathtt{5.76}}{\mathtt{\,\times\,}}{\mathtt{0.25}} \Rightarrow {\mathtt{h}} = {\mathtt{1.44}}$$

*Remember that the result of using these values in R(n) = h ^{n/2}/k will be in millions of dollars.*

For the sum of R(n) to n, notice that it can be written as:

sum(n) = (a + a^{2} + a^{3} + ... + a^{n})/k where a = h^{1/2}

The numerator is a geometric progression giving the result of sum(n) = a*(a^{n}-1)/[(a-1)k]

or sum(n) = (h^{1/2})*(h^{n/2}-1)/[(h^{1/2}-1)k}

See if you can now have a go at part b).

.

Alan Feb 23, 2015