seqser

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1

Name:_______________________ Date assigned:______________ Band:________

Precalculus | Packer Collegiate Institute

Boxes, Lottery Tickets, and Infinite Elephants, Oh my!

Section 1: Puzzles!

Puzzle #1

How many little squares are in the 42nd1 figure?

Generalize the result: How many little squares are in the nth figure? Extend the generalization: How many little squares are in the zero‐th figure?

Graph the result:

1 http://ind.pn/NfegPy

figure number

squares

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Puzzle #2

Part I: How many squares are in the 42nd figure? Generalize the result: How many squares are in the nth figure?

Part II: If the shaded area of the first figure is 81, what is the area of the 42nd figure? Generalize the result: What is the area of the nth figure?

Graph the results:

Graph the results:

figure number

squares

figure number

area

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Puzzle #3a

The number of small tiles in the nth figure is: If you had 12 tiles, the largest figure you could build would be the 3rd figure (you don’t have enough tiles to build the 4th figure). If you had exactly 7,570 tiles, the largest figure you could build would be the ___ figure. Explanation:

Puzzle #3b

The number of small tiles in the nth figure is: If you had 12 tiles, the largest figure you could build would be the 3rd figure (you don’t have enough tiles to build the 4th figure). If you had exactly 7,570 tiles, the largest figure you could build would be the ___ figure. Explanation:

Puzzle 3a: Puzzle 3b:

figure number

small squares

figure number

small squares

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Puzzle #4: Gardens are framed with a single row of border tiles as illustrated here

Draw the 4th garden:

Part I: How many border tiles are required for a garden of length 10?

Part II: How many border tiles are required for a garden of length 30?

Part III: How many border tiles are required for a garden of length 1000? Show and explain how you got your answer. Now that you’ve found the answer one way, come up with a second (different) way to “count” the border tiles for a garden of length 1000.

Part IV (generalize the result): If you know the garden length (call it n), explain how you can determine the number of border tiles.

Part V: Show how to find the length of the garden if 152 border tiles are used.

Part VI: Can there be a garden that uses 2012 tiles? What about 2013 tiles? Explain your reasoning.

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Part VII: Graph the results

figure number

border tiles

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Section 2: Mathematical Terminology

Each of the puzzles had you generate a set of numbers for the 1st figure, 2nd figure, 3rd figure, 4th figure, etc. In

mathematics, we call this a sequence.

For example, for Puzzle #1, you saw the pattern 1, 3, 5, 7, …

And we have notation for this. We’ll call this sequence { }nR (but we could just as well call it { }nBadger or { }nSnake ).

We use the superextrafancy curly brackets to indicate it’s a sequence, and we use the subscript to say where in the

sequence we are. So:

Instead of saying… the 5th number in this sequence R… we say 5R

Instead of saying… the 27th number in this sequence R… we say 27R

Instead of saying… the nth number in this sequence R… we say nR

As you’ve seen, the terms in a sequence can grow bigger or smaller, and we shall see that they can be crazy and get

bigger and smaller and bigger and smaller!2

Although there are a number of different kinds of sequences (as we shall see), we will really focus on two particular

kinds.

In Puzzle #1 and Puzzle #4, we saw the graphs look linear and the equation for the nth term was a linear equation. You

can now laugh, because we don’t call these sequences linear. We call them arithmetic. That’s because arithmetic is

about adding and subtracting, and for each term in the sequence we are adding and subtracting a fixed amount. The

hallmark of an arithmetic sequence is that there is a common difference between each term (if you subtract any term

from the previous term, you always get the same common difference).

In Puzzle #2, we saw the graphs look exponential and the equation for the nth term was an exponential equation. You

can now laugh again, because we don’t call these sequences exponential. We call them geometric, which has something

to do with the “geometric mean” (a geometry concept that I am going to ignore here). The hallmark of a geometric

sequence is that there is a common ratio between each term (if you divide any term by the previous term, you always

get the same common ratio).

2 Some sequences are tricky to figure out. Here’s a fun one:

1, 11, 21, 1211, 111221, 312211, . .. LookAndSay

Can you figure out 6

LookAndSay ? _____________________________________ (solution: http://bit.ly/KBeiSd)

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Section 3: Arithmetic Sequences

1. If you know the first number in an arithmetic sequence is 5 and each term in the sequence goes up by 3 , come

up with a formula for the nth term.

2. If you know the first number in an arithmetic sequence is 5 and each term in the sequence decreases by 3 , come up with a formula for the nth term.

3. If the first term in an arithmetic sequence is 1a and the common difference is d , what is the formula for na ?

4. If you know the seventieth number in an arithmetic sequence is 5 and each term in the sequence decreases by

3 , come up with a formula for the nth term. (Hint: your work for the previous problem will help you!)

5. If you know the fifth number in an arithmetic sequence is 5 and the eleventh number is 71 , come up with a

formula for the nth term.

6. If you know the fifth number in an arithmetic sequence is 5.2 and the eleventh number is 9.4 , come up with

a formula for the nth term.

Extra Practice

Arithmetic: Section 12.2 #3, 5, 7, 10, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33

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Section 4: Geometric Sequences

7. If you know the first number in a geometric sequence is 5 and the common ratio is 3 , come up with a formula

for the nth term.

8. If you know the first number in a geometric sequence is 5 and the common ratio is 1/ 3 , come up with a


9. If you know the first number in a geometric sequence is 1a and the common ratio is r , come up with a formula

for the nth term.

10. If you know the fifth number in a geometric sequence is 80 / 81 and the common ratio is 2 / 3 , come up with a

formula for the nth term. (Hint: your work for the previous problem will help you!)

11. If you know the third number in a geometric sequence is 54 and the fifth number is 486 , come up with a


12. If you know the fourth number in a geometric sequence is 156.25 and the ninth number is 488281.25 , come

up with a formula for the nth term.

Extra Practice:

Geometric: Section 12.3 #9, 11, 13, 15, 19, 21, 23, 25, 27, 29, 33, 35, 37

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Section 5: The Forwards Problem: Go From Formula to Sequence

Example: 1(

{1

} { })n

n ns

, so:

1s 2s 3s 4s 5s 6s 7s

1 1

2

1

3

1

4

1

5

1

6

1

7

Notice what is happening to this sequence as we go further and further along… although the numbers hop above and

below the x‐axis, we see that the terms are getting closer and closer to 0.

Will any of the dots ever lie on the x‐axis? How do you know? Convince me.

Geogebra Interlude

To make this graph, open Geogebra. In the input bar at the bottom type:

Sequence[(n,(‐1)^(n+1)/n),n,0,16]

What this does is it graphs the points 1( 1))( ,

n

nn

for n=0 to n=16. Be careful with the parentheses and watch out for

that extra “n” which I bolded.

To resize your window so you can see everything, click on the button at the top, and then place your arrow on the

y‐axis, click and hold down the button while drag the cursor up and down. The same goes for the x‐axis.

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Armed with basic geogebra knowledge, answer the following questions:

Given the following sequences, write out the first seven terms and then graph both in geogebra.

1. (a) 2 1

} }2

{ {n

n

na

(b)

2

2{ } { }

n

nbn

Use Geogebra to graph the first 16 values of these sequences.

What I entered in Geogebra for { }na : What I entered in Geogebra for { }nb

Sequence[ ] Sequence[ ]

Change your window to [0,16]x[0,10] Change your window to [0,16]x[0,250]

A rough sketch of what I see: A rough sketch of what I see:

1b 2b 3b 4b 5b 6b 7b

1a 2a 3a 4a 5a 6a 7a

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Section 6: The Backwards Problem: Go From Sequence to Formula

1. Given the first few terms of a sequence, can you come up with a formula that defines it? Is the sequence arithmetic,

geometric, or neither. Briefly explain how you decided your answer.

(a) 5, 6, 7, 8, 9 ... thus na

(circle one) arithmetic, geometric, or neither Explanation:

(b) 9, 16, 23, 30, ... thus2, na


(c) 9, 16, 23, 30, ... thus2, na


(d) 1 1 1 1

, , , , ... thus 3 9 27 81

1, na


WORK

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(e) 1 1 1

, , , , ... thus 3 27 81

11,

9 na


(f) 4 6 10

, , , , ... thus 9 7 5 3

2 8,

11 na


(g) .02, .002, .0002, .00002,... thus.2, na

(circle one) arithmetic, geometric, or neither Explanation: (h) (baby challenge)

2, 6, 24, 120, 720 ... thus1, na

(i) (challenge)

6, 12, 20, 30, 42, 56 ... thu2, s na

(j) (uber‐challenge)

1, 10, 33, 76, 145, 246, 385, 568, 801,

... thus

0,

na

(k) (ultra‐challenge)

1, 2, 3, 5, 8, 13, 21... thu1, s na

Hints for the challenges: (h) http://bit.ly/Mcg3FT … interesting!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! (i) it’s a quadratic (j) it’s a cubic

(k) the answer is 5) ((1 5)

2

1

5

n n

n na

. Weird, huh.

I guess that isn’t much of a hint as the answer. But isn’t it

strange that even though the formula involves 5 , you

always get an integer output.

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Section 7: An Introduction to Arithmetic Series A prelude (from http://bit.ly/MC7YHk) About 100 years ago, a young boy (who grew up to be a great mathematician) by the name of Gauss (pronounced "Gowss") was at school when the class got in trouble for being too loud and misbehaving. Their teacher, looking for something to keep them quiet for a while, told her students that she wanted them to "add up all of the numbers from 1 to 100 and put the answer on her desk." She figured that would keep them busy for an hour or so. About 30 seconds later, the 10‐year‐old Gauss tossed his slate (small chalkboard) onto the teacher's desk with the answer "5050" written on it and said to her in a snotty tone, "There it is." Let us look at the following diagram. We can come up with a sequence for the number of boxes in each figure.

The sequence is 1, 3, 6, 10, 15, 21, …

However, if we want to find the nth term in the sequence, we have a problem. It turns out (and we’ll show this) that the

formula is: 21 1

2 2n ns n … or written more elegantly, ( 1)

2n

n ns

.

WHAT IN THE WHAT? How in the world does that work?

1. Compare each figure to the previous one. Describe how the nth figure is changing based on the n‐1th figure.

If we want the number of squares in the nth figure, we have to add together a bunch of numbers.

For the fifth figure, we add 5 1 2 3 4 5s

For the ninth figure, we add 9 1 2 3 4 5 6 7 8 9s

For the nth figure, we add 1 2 3 ... ( 2) ( 1)ns n n n

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2. But if we want to find what this sum is, we are going to have to add together a lot of numbers. Which is

annoying. Here’s a shortcut. Let’s calculate 5s in a special way, that might seem convoluted. We’ll add the sum

to itself, but in a special way.

5

5

52 6

1 2 3 4 5

5 3

6

1

6

2

6 6

4

s

s

s

Now we see that 5 6( 302 5)s . Thus 5 15s . Which we know.

Check yo’self! Using this method, find 10s .

Practice one more time. Be Gauss. Find the sum of the first 100 positive integers: 100s .

3. Now try it more generally for 1 2 3 ... ( 2) ( 1)ns n n n

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4. Does this work for other sequences? Try this technique out with:

7, 10, 13, 16,4, 19...

Find the sum of the first five numbers by adding them: __________

Find the sum of the first five numbers by using the technique… Does the technique work? (If it doesn’t, explain

why not.)

5. Try this technique out with:

3, 8, 13, 18,2 23, ...,

Find the sum of the first six numbers by adding them: __________

Find the sum of the first six numbers by using the technique… Does the technique work? (If it doesn’t, explain

why not.)

6. Try this technique out with:

4, 8, 16, 32, 62, 4, ...

Find the sum of the first five numbers by adding them: __________

Find the sum of the first five numbers by using the technique… Does the technique work? (If it doesn’t, explain

why not.)

Key Mathematical Conclusion: This technique of adding the sum to the original sum, but reversing the order of

the terms, works for __________________________ series because___________________________________

__________________________________________________________________________________________.

It will not work for ___________________________ series because ___________________________________

__________________________________________________________________________________________.

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7. Generalize things now! To find the sum of an arithmetic sequence, you need to know the first term, the last

term, and the number of terms total. Write an equation (using only the terms “first term” “last term” and

“number of terms”) which gives you the sum.

Sum of an Arithmetic Series

8. A proof without words.

Yeah, it may be a “proof without words,” but you need to words to explain that you understand it. Explain how this

“proof without words” is a visual illustration of the equation you came up with in the previous problem.

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Section 8: Sigma Terminology and Notation

So we’ve talked about adding the integers from 1 to 100 together. There is a mathematical way to say that. It looks fancy

and possibly scary, but it isn’t! Promise.

100

1

1 2 3 ... 98 99 100i

i

If you don’t understand this, let me show you a few other examples of our fancy notation in action:

1

26

1 4 9 16 25 36n

n

7

2

(2 5) 1 1 3 5 7 9p

p

5 6 7 88

5

1 1 1 1 12 2 2 2

3 3 3 32

3k

k

12

1 1 1 1 1 1 1

1 4 9 16 25 36...

n n

(weird fact3)

The variable itself is just a placeholder… any letter will do! Just make sure you pay attention to the top and bottom

numbers!

1. Represent the following sums using sigma notation:

(a) (problem 4 from the previous section): 4 7 10 13 16

(b) (problem 5 from the previous section): 2 3 8 13 18

(c) (problem 6 from the previous section): 2 4 8 16 32

3 Okay, here’s a huge surprise. If you add all these terms up, the sum will get closer and close to

2 / 6 . WHAAAAT? WHY IS PI INVOLVED IN THIS AT ALL?! I know, so very strange. Is it related to circles? Calculus can help you understand this here. I know, I know, you’ll have to wait a bit. Also, this series is tied up with something called the Riemann‐Zeta function. You might not have heard about it, but understanding the zeros to this function will literally make you a millionaire. Check out the million dollar problems (including the Riemann Hypothesis) here: http://bit.ly/LX4nHv

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2. Expand the sigma notation to show the sum. You do not need to actually find the sum:

(a) 4

1

5 3( 1)i

i

(b) 5

1

29

3

k

k

(c) 1

4

(10)ii

(d) 9

3

2

2i

i

i

(e) 2

9

3 4

2

n

n

n

3. Now we’re going to cycle back to arithmetic series. First, look at the following problems below and before you

find the sum, explain (in words) how you know these are arithmetic series (as opposed to geometric, or

something else).

Explanation:

(a) 10

5

2 3i

i

(b) 10

5

2 3( 1)i

i

(c) 15

1

5 2i

i

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(d) 6

6

1 67i

i

(e) 100

23

1 67i

i

Additional Problems:

Sigma Notation:

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Section 9: An Introduction to Finite Geometric Series!

Lotto! Money for Life!

You are going to see if you’re going to win a million dollars! Below are three scratch offs – but someone has already

scratched off the first two circles. If the three scratch offs all show $ under them, you win a million dollars that will be

paid to you in $50,000 installments at the end of each year for 20 years. If you see a cherry, you win a piece of candy. If

you see anything else, you win nothing.

Some of you won, some of you lost. For us, here, now, in math class, let’s assume you won, and you want to maximize

your money in safe way, so when you’re 36 or 37 you have a pile of money that you are sitting on.4 You have a ton of

patience, so you have this money direct deposited in a bank account which gives you 2% interest, earned at the

beginning of the year. Let’s check to see how much money your bank account will show at the end of the each year.

1. How much money do you have at the end of the first year?

Answer: You have $50,000. This is because you haven’t yet earned interest on this (interest is earned at the

beginning of the following year.)

2. How much money do you have at the end of the second year?

3. How much money do you have at the end of the third year?

4. How much money do you have at the end of the fourth year?

5. Can you write your answer to #4 using summation notation?

6. Can you write how much money you’ll have at the end of 20 years using summation notation?

4 We are going to ignore taxes for now. However, they could be factored in with a little effort.

$ $

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7. Do you see that each term in the sum above forms a geometric sequence?

The first term is _______________ and the common ratio is _______________.

What we have is a geometric series! And we saw that the technique to sum an arithmetic series doesn’t work for

geometric series.

I’m going to show you a technique to add a geometric series! Let’s consider a four‐term series with first term of

5 and a common ratio of 4.

2 35 5(4) 5(4) 5(4)s

Let’s multiply s by the common ratio to get 4s .

2 3 44 5(4) 5(4) 5(4) 5(4)s

And now let’s subtract the two equations!

2 3 4

2 3

4

5 5(

4 5(4) 5(4) 5(4) 5(4)

5(4)4) 5 (4

3 5(4)

)

5s

s

s

Thus we have 45(4)

4255

3s

Will this technique always work? Try it out by calculating how much money you’ll have at the end of the fourth

year! See if the algebra works out. And then compare your answer to your sum on the previous page.

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8. Use this technique to calculate how much money you’ll have at the end of the twentieth year! Because of

interest, it should be more than a million dollars. How much more money than a million dollars have you made?

Practice!

9. Add the first seven terms of the series in this manner 5 5 5 5 5 5

52 4 8 16 32 64

s

10. If you wrote out ten terms, what would be the tenth term in this sum? What about the fifteenth term? What

about the fiftieth term? What about the nth term?

11. Using this new technique, exactly find the sum of the first fifteen terms.

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12. (a) Using this new technique, exactly find the sum of the first n terms.

(b) As n gets bigger and bigger, what happens to the sum?

13. (a) If the series were altered, so that it is: 1 1 1 1

3 9 27 81...s , exactly find the sum of the first n terms.


14. (a) If the series were altered, so that it is: 23 23 23 23

3 9 27 8.

1..s , exactly find the sum of the first n terms.


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15. If the series were altered, so that it is: 3 3 3 6 12

4 10 25 125 625...s

(a) Explain how you know this series is a geometric series.

(b) Exactly find the sum of the first eight terms. Write the sigma notation for the sum of the first 8 terms.

(c) What is the nth term in this series?

(d) Exactly find the sum of the first n terms. Write the sigma notation for the sum of the first n terms.

(e) As n gets bigger and bigger, what happens to the sum?

16. If the series were altered, so that it is: 1 2 4 8s (it is finite and doesn’t go on forever!), use this

technique to find the sum of these four terms. Then check your answer by adding these four terms together.

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17. Does infinity equal ‐1?

Using your brain, what is the sum of this infinite geometric series: 1 2 4 8 16 ...s

Now let’s use the technique we’ve perfected above.

1 2 4 8 16 ...s

2 2 4 8 16 32 ...s

1s

Thus we can see that 1s .

Explanation the discrepancy between your brain answer and our procedural answer. Which is the correct sum?

18. If the series were altered, so that it is: 2 3 4 ...s ar ara ar ar , we are designating the first term as a

and the common ratio as r .

(a) Explain how you know this series is a geometric series.

(b) Exactly find the sum of the first eight terms. Write the sigma notation for the sum of the first 8 terms.

(c) What is the nth term in this series?

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(d) Exactly find the sum of the first n terms. Write the sigma notation for the sum of the first n terms.

(e) As n gets bigger and bigger, what happens to the sum?

Key Mathematical Conclusion: This technique is powerful and can be used to find the sum of the first n terms of

________________ sequences. The reason this technique works is because _____________________________

___________________________________________________________________________________________

___________________________________________________________________________________________

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Section 10: Infinite Geometric Series!

Infinity!

0. First, watch this: http://bit.ly/NbitTN

1. I am giving you 8 infinite geometric series. Add the first 20 terms using the formula you came up with! Write

your answer next to the series…

(a) 1

1

2

n

n

(e) 1

1

1

104

n

n

(b) 1

1

1

23

n

n

(f) 1

1n

n

(c) 1

2n

n

(g) 1

2

35

n

n

(d) 1

1.01n

n

(h) 1

1

2

35

n

n

2. Put an * next to the ones you think will go off to infinity if you keep on adding all the remaining terms!

3. Explain why (f) should definitely have an * next to it.

4. Explain why you chose to give or not give (c) an asterisk.

5. Explain why you chose to give or not give (b) an asterisk.

6. Explain why you chose to give or not give (d) an asterisk.

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7. Proof without words.

Yeah, it may be a “proof without words,” but you need to words to

explain that you understand it. Explain what this “proof without

words” is a visual illustration of. (Hint: The large square is a 1 by 1

square.)

8. Now we are going to consider geometric series with a negative common ratio! Add the first 20, 21, 22, and 23

terms using the formula you came up with! Use these sums (we call them partial sums) to conjecture whether

the infinite series is convergent or divergent (put an * next to the ones you think are divergent).

(i) 1

1

2

k

k

20 terms: 21 terms: 22 terms: 23 terms:

(j) 1

1

104

n

n

20 terms: 21 terms:

22 terms: 23 terms:

(k) 1

31

2

i

i


(l) 1

1p

p

20 terms: 21 terms:

22 terms: 23 terms:

(m) 1

2m

m


(n) 1

52

3

n

n

20 terms: 21 terms:

22 terms: 23 terms:

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(o) 1

1.01k

k


(p) 1

1

52

3n

n

20 terms: 21 terms:

22 terms: 23 terms:

Key Mathematical Conclusion: Infinite geometric series will shoot out to infinity if ___________________________

______________________________________________________________________________________________.

Mathematically we call this sort of series divergent. However infinite geometric series will get closer and closer and

closer to particular (finite) number if ________________________________________________________________.

We call this sort of series convergent.

If the series is convergent, you can figure out what the sum is approaching! In problem 18 of the previous section,

you determined that the sum of the first n terms of a geometric series is:

As n increases to infinity, we can say that one term in that equation becomes negligible. What term is that, and

why?

As a result, we can determine that the sum, as we have more and more terms, approaches:

Decide if each of these series are convergent or divergent. If they are convergent, write down what number the

series converges to next to the sum.

(a) 1

1(3)

2n

n

convergent / divergent

(d) 1

1( 1)

10n

n


(b) 1

1

23

n

n


(e) 1

(1.7210 )0 n

n


(c) 1

3

25

n

n


(f) 1

(0.7210 )0 n

n


seqser

Documents

0 nn convergentdivergent

sequence

convergentdivergent

20terms

na

arithmetic

ifyouhadexactly7

comeupwithaformulaforthenthterm