Algebra II / Mr. Hansen 
Name: _________________________________ 
Preparation for Quiz Friday 4/23/99 covering Sections 115 through 117
Solutions to "Do Slowly" and "Regular" HW from §117
(note: these include ideas from §115, §116, and §117)
S1. 
t_{20} = 34(1.7)^{19} = 812,846.2813 
S2. 
t_{30} = 78+29(3.2) = 170.8 
S3. 
You may be able to see that d = 11 without any work. Consequently, t_{2} = 55. 
S4. 
We can compute r by dividing any term by the one just before it. We are given two terms, t_{1} and t_{2}, so divide t_{2} by t_{1} to get 80/100 = 0.8 = r. Therefore, we have these answers: 
S5. 
This problem is similar to S4, except that r = 0.95 instead of 0.8. It's a geometric sequence (note: we don't say "series" this time since there's no discussion of partial or infinite sums). By the formula at the bottom of p.567, t_{10} = 800(0.95)^{9} » 504.1995278. 
S6. 

S7. 

S8. 

S9. 

S10. 

1... 

a. 

b. 

c. 

3.a. 

b. 

c. 
Solve the inequality t_{n} £ 0.25 for n. Can you do this? By the formula at bottom of p.567, t_{n} = t_{1}(r^{n}1) = 0.96(0.96)^{n}1 = 0.96^{n}, and this final expression must be £ 0.25. Take logs of both sides of our new inequality (namely 0.96^{n} £ 0.25) to get n log 0.96 £ log 0.25, or n ³ (log 0.25)/(log 0.96). By the way, why does the direction of the inequality flip when we divide both sides by log 0.96? This final expression is approximately 33.96, and since n must be the next integer that is ³ 33.96, we take n = 34 washings. 
d. 
An arithmetic sequence (with fixed d) would predict that the fraction of color remaining keeps decreasing at a constant linear rate. But this makes no sense; the fraction can't be negative, as this model would eventually predict. 
4... 
Let t_{1} = distance traveled between 1st and 2nd bounce, let t_{2} = distance traveled between 2nd and 3rd bounce, etc. In other words, we will let t_{n} = distance traveled between the nth and (n+1)st bounce. 
a. 
t_{1} = 36 m (reason: the height is 90% of 20 m, or 18 m, and the ball must travel up to that height and then back down) 
b. 
Each bounce is 90%, 0.9, as high as the one before. Thus we have a geometric sequence with r = 0.9. 
c. 
t_{6} = 36(0.9)^{5} = 21.25764 » 21.26 m 
d. 

6... 

a. 
t_{365} = 0.105(1.05)^{364} » $5,421,184.16 
b. 
t_{366} = 0.105(1.05)^{365} » $5,692,243.37, which is $271,059.21 more for a single day! 
14... 
Let t_{1} = amount the parents pay on day 1, let t_{2} = amount the parents pay on day 2, etc. We can see that this is a geometric sequence with t_{1} = 0.01 and r = 2. After 30 days, those poor, gullible, misguided parents will have paid S_{30} = 0.01(1r^{n})/(1r) = 0.01(12^{30})/(12) = 0.01(1,073,741,823)/(1) = $10,737,418.23. Warning: If you get your parents to go for this idea, I cannot assume any responsibility! 
26... 
First, note that 4% annual interest can be written as 0.04. Divide by 12 to get the interest rate in each month, namely 0.04/12. Add this to 100% (i.e., 1) to get the total factor that the account balance is multiplied by each month. (There's 100% of the money, plus 4%/12 for the monthly interest.) This factor, namely, 1.0033333, is r. In this problem, the rest of the calculations should be performed using months, not years. 
a. 
After 1 month, we have $20,000(1.0033333), or $20,066.67, and this is our t_{1}. From this we can get the amount after 10 years (i.e., 120 months), which is t_{120} = 20,066.67(1.0033333)^{119} » $29,816.65, and the amount after 50 years (i.e., 600 months), which is t_{600} = 20,066.67(1.0033333)^{599} » $147,290.43. 
b. 
After 1 month, we have $100(1.0033333), or $100.33, and this is the first term of a new sequence, or actually we should say series since we'll be adding up all these $100 payments with their interest. Let u_{1} = balance at end of 1st month, u_{2} = balance at end of 2nd month, etc. We use u for the terms, instead of t, to avoid confusion with the sequence in part (a). After 10 years (120 months), our balance using the method of $100 monthly payments is S_{120} = 100.33333(1r^{n})/(1r) = 100.33333(11.0033333^{120})/(11.0033333) » 100.33333(0.4908267)/(0.0033333) » $14,774 to the nearest dollar, but since another $100 payment (without interest) arrives on the first of the month, we have $14,874 total after 10 years. After 50 years (600 months), our balance is S_{600} = 100.33333(11.0033333^{600})/(11.0033333) » 100.33333(6.364375)/(0.0033333) » $191,572 to the nearest dollar, but again we must add $100 for the newly arriving payment to get $191,672 total after 50 years. 
c. 
You could solve an inequality for n, but that is rather tedious. You could also use the sequence mode of your TI83, but that is even worse, since you have 600 data points to plot. Instead, I recommend using your TI83 and pretending that the expressions for t_{n} in part (a) and S_{n} in part (b) are really continuous functions of x instead of discrete functions of n. Begin by "STO"ing 1+.04/12 into R so that R is 1.0033333. (This makes sense, because you're keeping 100%, or simply 1, plus a monthly percentage that is 1/12 of the annual percentage.) Since t_{1} is 20000*R, we can write t_{n} as t_{1}r^{n}1 = 20000R(R^(N1)) = 20000R^N, or using our "function of x" trick, Y_{1} = 20000R^X. Actually press your "Y=" key and punch this function in, please. Similarly, punch in Y_{2} = 100R(1R^X)/(1R)+100. (Why does Y_{2} have a "+100" at the end?) Now graph with a window of [0, 600] × [0, 200000]. You'll see that the second curve overtakes the first curve about midway through the 50year (600month) span. Use 2nd CALC to find the actual intersection point, namely X=__________. Therefore, the second curve will exceed the first curve after 328 months (27.25 years). If you think you will live that long, you should choose the $100permonth payout instead of the lump sum. Note: Strictly speaking, you should not rely on smooth curves, since the domain for both functions is {natural numbers}, not Â . However, the smooth curves are much easier to work with than sequences in this case. In a problem involving a smaller domain (say, the integers from 1 to 25), the TI83 "Seq" mode would be more reasonable. However, you have to define partial sums in a special way on your calculator, and we probably won't have time to discuss this in class. (For advanced students: Use a recursive sequence definition to handle the partial sums.) 
d. 
Change R by "STO"ing 1+.07/12 into R. Then do a ZOOM ZoomFit to redraw the curves. Can you see that this time, the first curve starts higher and increases faster than the second curve? In other words, if the assumed interest rate is 7% instead of 4%, the $100permonth payout plan can never catch up with the lump sum plan. For advanced students: One can show (via tricky algebra) that an annual interest rate of 5.6745% or more prevents the $100permonth payout plan from catching up with the lump sum plan in the first 50 years. By using some even trickier algebra and taking the limit as x (or n) grows without bound, one can show that an interest rate of 6.0302% or more prevents the $100permonth payout plan from ever catching up with the lump sum plan. 
e. 
If you consider the effects of annual income tax paid from the proceeds of the account, you have a lower net interest rate. For example, if you live in Virginia and have a federal tax rate of 28% and a state tax rate of 5.75%, you have to pay more than a third of your interest earnings as tax, and if the nominal rate is 4%, your actual aftertax rate is only 2.65%. Since lower interest rates make it harder for the lump sum plan to maintain its lead, the effect of taxes could very well change the conclusions we reached in parts (c) and (d). If you really want to be sophisticated, you could compare the two payout plans under various tax deferral or tax avoidance strategies, such as a 401(k) plan or a Roth IRA. 