Once a week Mrs. Baker makes sugar cookies. The first week she makes the recipe, she uses the full 2 cups of sugar called for. Each week after that, she reduces the amount of sugar by one third.
(a) How much sugar does she use for the cookies on the fifth week?
(b) How much sugar does she use for cookies over half a year?
(c) If Mrs. Baker became immortal and baked cookies every week for all eternity, how much sugar would she use?
If Mrs. Baker reduces the amount of sugar by each week, that means she keeps of the previous amount of sugar. So on the second week she uses cups of sugar, and so on. Let's make a table.
(a) From the table, we can see that on the fifth week Mrs. Baker uses
cups of sugar.
(b) The amount of sugar Mrs. Baker uses each week is described by a geometric series with a = 2 and . We want to know how much sugar she uses over half a year, or 26 weeks. This means we need to take the partial sum of the geometric series with n = 26. We get
She uses almost 6 cups of sugar.
(c) To find the amount of sugar Mrs. Baker uses if she bakes cookies every week for all eternity, we need to find the infinite sum of the geometric series:
Mrs. Baker uses 6 cups of sugar if she bakes cookies every week for all eternity, and from (b) we know that almost all that sugar is used in the first half year of baking.
Mr. Vold is a sadistic teacher who likes writing lots of exam questions. He usually starts out the semester with only 10 questions on the first exam, but for each subsequent exam he writes one and a half as many questions as were on the previous exam! Since there's no such thing as half a question and Mr. Vold likes writing questions, round your answers up to the next integer.
(a) How many questions are on the second exam of the semester?
(b) How many questions are on the third exam of the semester?
(c) How many questions are on the fifth exam of the semester?
(d) If Mr. Vold wrote 20 exams in a semester, how many total exam questions would they have all together?
The question says for each exam Mr. Vold writes "half again as many questions as were on the previous exam."
This means the number of questions for an exam is
(a) The first exam had 10 questions. The second exam has
(b) The third exam has
which rounds up to 23 questions.
(c) There are (at least) two ways to interpret this question, depending on when we deal with rounding.
In part (b) we found that there were 23 questions on the third exam. In this case there are
which rounds up to 35 questions on the fourth exam.
Then there are
which rounds up to 53 questions on the fifth exam.
We could leave the rounding until the end, instead. If the first exam has 10 questions then the second exam has
the third exam has
and the fifth exam has
This rounds up to 51 questions on the fifth exam.
(d) For this question we don't want to deal with partial rounding. We're looking at a geometric series with a = 10 and . To find how many questions Mr. Vold writes over 20 exams we need to add up the first 20 terms of this geometric series:
Rounding up, we see Mr. Vold will write 66486 questions over the semester.
Mo reads a lot of books. When he was five years old he read 4 books. Each year he reads three more than twice the number of books he read the previous year.
(a) How many books does Mo read when he's six years old?
(b) How many books does Mo read when he's eight years old?
(c) Give a formula for the number of books Mo reads when he's 5 + i years old. It's ok if there's a summation sign in your formula.
(hint: When finding the answers to (a) and (b), don't simplify too much).
(d) Use your formula from (c) to calculate how many books Mo reads when he's fifteen years old.
Check your answer by computing the same number without using the formula.
For n ≥ 5, let
an be the number of books Mo reads when he's n years old. We're told
a5 = 4
to start off, and that to get from one number to the next we need to multiply by 2 and then add 3.
This means for n > 5,
an + 1 = 2an + 3.
(a) Take a5, multiply by 2, then add 3:
a6 = 2a5 + 3
= 2(4) + 3
Mo reads 11 books when he's 6 years old.
(b) When Mo is seven he reads
a7 = 2(11) + 3 = 25 books.
When he's eight he reads
a8 = 2(25) + 3 = 53 books.
(c) Since we're asked to give a formula for the number of books Mo reads when he's 5 + i years old, we'll write a6 as a5 + 1,
a7 as a5 + 2, and so on.
Following the hint, look at how we got the answers to (a) and (b) without simplifying so much.
The first one was simple:
a5 + 1 = 2(4) + 3
Instead of simplifying, put this answer straight into the equation for a7 = a5 + 2:
a5 + 2 = 2a5 + 1 + 3
= 2(2(4) + 3) + 3
= 22(4) + 2 × 3 + 3
Now we can put the non-simplified expression
a5 + 2 = 22(4) + 2 × 3 + 3
into the equation for a5 + 3:
a5 + 3 = 2a5 + 2 + 3
= 2(22(4) + 2 × 3 + 3) + 3
= 23(4) + 22 × 3 + 2 × 3 + 3
Writing all our expressions next to each other, we have
a5 + 0 = 4
a5 + 1 = 2(4) + 3
a5 + 2 = 22(4) + 2 × 3 + 3
a5 + 3 = 23(4) + 22 × 3 + 2 × 3 + 3
We can generalize to get an expression for a5 + i:
This expression has 2 parts. The first part is 4 multiplied by a power of 2. The second part is itself a series. Writing the series in summation notation, we can rewrite the expression as
Don't be confused by the use of j in the summation. We need some letter for the index of summation, and n and i have already been used in this problem.
(d) We want to find
a15 = a5 + 10.
Using the formula from (c),
The second piece of the expression is a geometric series with first term
a = 20 × 3 = 3
and ratio 2. There are 10 terms from j = 0 to j = 9, so take n = 10. This partial sum is
Adding up the pieces of the expression,
a15 = 4096 + 3069 = 7165.
Check the answer by direct computation. We left off before at a8 = 53, so
a9 = 2(53) + 3 = 109
a10 = 2(109) + 3 = 221
a11 = 2(221) + 3 = 445
a12 = 2(445) + 3 = 893
a13 = 2(893) + 3 = 1789
a14 = 2(1789) + 3 = 3581
a15 = 2(3581) + 3 = 7165
Thankfully, this answer agrees with what we got using the formula. We conclude that Mo reads an absolutely ludicrous number of books the year he's 15: 7165 books.
On January 1st Komi puts $100 into his bank account. On the first of each month after that he deposits an additional $10.
(a) How much money is in Komi's account at the end of February?
(b) How much money is in Komi's account at the end of March?
(c) How much money is in Komi's account after n months?
Here's the timeline for the money going into Komi's account:
(a) By the end of February Komi has made deposits of $100 (in January) and $10 (in February) so his account contains $110.
(b) By the end of March Komi's account contains $10 more than it did at the end of February, for a total of $120.
(c) At the end of one month (end of January), Komi's account contains $100. At the end of two months (end of February), the account contains
100 + 10.
At the end of three months (end of March), the account contains
100 + 10 × 2.
Generalizing, at the end of n months Komi's account contains
100 + 10(n – 1) dollars.
And in case you were wondering, no, this problem doesn't have much to do with arithmetic or geometric series. You could look at the amount of money in Komi's account as $100 plus a partial sum of an arithmetic series with d = 0, but why bother?
On January 1st Kendra puts $100 into her bank account. On the first of each month after that she deposits an additional $100. Unfortunately, during the middle of the month she always has to withdraw 75% of her money to pay bills.
(a) How much money is in Kendra's account at the end of March?
(b) How much money is in Kendra's account at the end of n months? Write your answer in closed form.
(a) Here's the timeline:
At the beginning of March, Kendra has $131.25. In the middle of March she withdraws 75% of that amount, leaving 25% in the bank.
Rounded to the nearest cent, her bank account holds
0.25(131.25) = $32.81
at the end of March.
(b) This question is similar to (b) of the example, but it asks about what's happening at a different point in the month.
At the end of the month, Kendra has 25% of the money she had at the beginning of the month. We'll call the amount she has left at the end of the month Ln. There are two ways to do this problem: we could use the answer to (b) of the example, or we could start from scratch. It's a good idea to understand both ways.
Using the example:
At the beginning of month n, Kendra has Mn dollars in her account.
At the end of month n, she has 25% of that amount left.
Starting from scratch:
Let Ln be the amount of money in Kendra's account at the end of n months. To get from Ln – 1 to Ln she deposits $100, then keeps 25% of the resulting amount.
Ln = 0.25(100 + Ln – 1 ).
This tells us how to build each Ln from the previous one, so look at some values of Ln.
The value Ln is the sum of a geometric series with r = 0.25,
a = 0.25(100) = 25,
and n terms (since the exponents go from 1 to n). So
Thankfully, this is equivalent to the formula we found when we used the example instead of starting from scratch.
Lizette decides that starting in January she will deposit $50 into her bank account at the start of each month. Her account earns 0.25% interest per month. Interest is calculated at the end of each month. Truncate answers to two decimal places.
(a) In the middle of February, how much money is in Lizette's account?
(b) In the middle of March, how much money is in Lizette's account?
(c) In the middle of the nth month (where January is the 1st month, February is the 2nd month, etc.), how much money is in Lizette's account? Give your answer in closed form.
(d) In the middle of December, how much money is in Lizette's account?
This exercise is very similar to the previous one. In January, Lizette deposits $50 and earns 0.25% interest. After the interest calculation, she has
1.0025(50) = 50.125
which we truncate to $50.12.
Remember that to calculate interest we multiply by (1 + ( interest rate)) since we have to account for the initial amount of money as well as the interest earned.
(a) At the beginning of February, Lizette adds $50 to the $50.12 she had from January:
Since interest isn't calculated until the end of the month, in the middle of February she still has
50 + 50.12 = 100.12.
(b) Extend the timeline one more month:
Lizette earns interest on her $100.12, then deposits another $50. In the middle of March she has
50 + 1.0025(100.12) = 150.37.
(c) Let Mn be the amount of money in Lizette's account in the middle of the nth month. Then Mn – 1 was the amount of money in her account in the middle of the (n – 1)st month.
To find Mn we calculate interest on Mn – 1, then add another $50. So
Mn = 50 + 1.0025Mn – 1.
Here are the values of Mn for some small n:
Since we're so good at recognizing these patterns by now, we see that Mn is the sum of a finite geometric series with a = 50,
r = 1.0025, and n terms:
(d) There are two ways to do this problem. We could use the formula from (c) to find M12, or we could calculate Mn for n = 1, n = 2, all the way up to n = 12. Since n = 12 is pretty small and we want to make sure we get the right answer, we'll do it both ways.
Using the formula from (c):
All we have to do is plug in n = 12 to the formula we found in (c). We get
which, since we're truncating our answers, works out to $608.31 in the bank.
Using direct calculation:
We found that, in March, Lizette had
M3 = 150.37.
Now we apply the formula
Mn = 50 + 1.0025Mn – 1
M4 = 200.74
M5 = 251.24
M6 = 301.86
M7 = 352.61
M8 = 403.49
M9 = 454.49
M10 = 505.62
M11 = 556.88
M12 = 608.27
We're 4 cents off from the number we got using the formula, because here we truncated each value of Mn along the way. The answers are close enough for us to be confident that the formula is correct.