Series 1 - Homework

[user1454]

Hello, could you please explain the solution for all these 3 questions below thoroughly?

In the first question, I approached the problem as follows:

1. We have a sequence: n , n+2 , n+4
2. This can be expresed as 600n + (2+4+ …)
3. I set n=3, which gives us : 1800 + (300*1202)
4. However, the answer is 361,200 , which suggests that n is considered to be 1 !
   But since the sequence starts from 3, n should indeed be 3!

Regarding the last question, I reviewed the solution and tried substituting the options into the non-equation to see which ones fit. For example, when I tested 239, it worked because it satisfies the condition > 0. So, I’m wondering why the only accepted answer is 279?

[cylverixxx]

it’s 2 + 4 + 6 + ... 1198, but i think you added 2000 as well.

After that yeah n = 3 clearly should work.

This is just the partial sum of a geometric sequence with common ratio \left (\frac 32\right)

For the last question, 239 doesn’t work because the sum on the left hand side is exactly 7260. 39,40, and 60 are too small by inspection. 240 is a multiple of 4 whereas the series isn’t, so yeah the only choice left to pick is 279.

[user1454]

Thank you for your reply.

In the first question, I calculated 600n + (300* 1202). I consider 1202 as the sum of 2 and 1200 , since the 600th term should be 2*600 = 1200 . Therefore, we are have 300 pairs of 1202 . Adding 1800 gives us 362,400 , which is incorrect . However, I couldn’t understand your response because it suggests the answer should be 361,800 , which is also incorrect. The correct answer is 361,200.

Could you please explain the second question in more detail? There weren’t any examples in the related learning videos, and I would appreciate a clearer explanation.

Regarding the last question, please check the solution for the part : 2n^2 + n - 7260 > 0 . When I substitute 239, it obviously satisfies the inequality! How did you conclude that it equals 7260? Is there a more logical way to understand this type of question? If so, I would be grateful for an explanation, as it still seems unclear to me.

Thanks you!

[cylverixxx]

@user1454

I consider 1202 as the sum of 2 and 1200 ,

Yeah my point is that the sum is actually 600n + (0 + 2 + 4 + \ldots + 1198), but you are instead doing 600n + (2 + 4 + \ldots + 2000).

For reference, 0 + 2 + 4 + \ldots 1198 = 359400. That added to 600(3) gives you 361200

@user1454

Could you please explain the second question in more detail? There weren’t any examples in the related learning videos, and I would appreciate a clearer explanation.

Maybe it’s helpful for you to watch a video on geometric series for context. Once you’ve watched a video then you can come and try to apply it on this question. If you, by chance, still don’t get it then i’m happy to help. It’s just pretty hard to teach you a whole topic on text.

@user1454

Regarding the last question, please check the solution for the part : 2n^2 + n - 7260 > 0 . When I substitute 239, it obviously satisfies the inequality! How did you conclude that it equals 7260? Is there a more logical way to understand this type of question? If so, I would be grateful for an explanation, as it still seems unclear to me.

No the sum to 239 is exactly 7260.

2n^2 + n - 7260 > 0 gives n > 60 \implies 4(n) - 1 > 239. I mean you can just plug n = 60 into 2n^2 + n - 7260 and see that it’s clearly 0.

@user1454

2n^2 + n - 7260 > 0 . When I substitute 239, it obviously satisfies the inequality!

Nah, the nth term of the sequence is given as 4n - 1. When n = 60 then you get 239, so what you should be plugging in is n = 60 not n = 239

[Leaderboard]

@user1454

Could you please explain the second question in more detail? There weren’t any examples in the related learning videos, and I would appreciate a clearer explanation.

Heed the note “the previous question may help”.

[user1454]

Hello,

Thank you for your quick responses!

I feel I have grasped a lot of the content, but I would like some clarification. Could you please explain how the expression “4n-1” was derived from the explanation in the solution to the last question?

This explanation:

Recall from the PrepSwift Arithmetic section that to find the sum of multiples in an interval, we

• find the number of multiples,
• and multiply by the sum of each “pair” in the series.

[cylverixxx]

@user1454

I feel I have grasped a lot of the content, but I would like some clarification. Could you please explain how the expression “4n-1” was derived from the explanation in the solution to the last question?

It’s an arithmetic sequence. If you haven’t learnt that yet then it’s a “sequence of numbers” with the same common difference between its succeeding and preceding terms.

Since the first difference is constant between consecutive terms then you can “extrapolate” a linear function to match the nth term. For example, assume the sequence takes the form y = an + b where a,b are coefficients to be found. You know (n,y) = (1,3) and (n,y) = (2,7) so you can uniquely find the values of a and b, which happen to be (a,b) = (4,-1). You can use this idea whenever you have a constant first difference (so any kind of sequence like this).

Otherwise, if you’re already familiar with arithmetic sequences then you have: a_n = a_1 + (n-1)d where (a_1, d) = (3, 4). Simplifying should yield a_n = 4n - 1.

Just as an extra “bonus” lesson for you, if the first difference wasn’t “constant” but the second difference was, then you’d be looking to extrapolate a “quadratic function” for your nth term. The idea generalizes, so basically if nth difference is constant then you can “extrapolate” an nth degree polynomial.

@user1454

This explanation:

Recall from the PrepSwift Arithmetic section that to find the sum of multiples in an interval, we

• find the number of multiples,
• and multiply by the sum of each “pair” in the series.

Yeah okay, I watched this for you and u can apply this just as you would in the video. The only slight difference is that our sequence is multiples of 4 shifted by 1 unit. You can imagine transforming the sequence (3,7,11, \ldots, 239) to (4,8, 12, \ldots, 240) and then just follow the same procedure. The transformation above helps us to count the number of pairs easily because we know there are 60 multiplies of 4 in the closed interval [4,240] thus giving you 30 pairs. It’s obvious that the number of pairs in both sequences must be the same cuz all we did is add 1 to each element (did not add or remove elements). Moreover, each pair in the sequence (3,7,11, \ldots, 239) sums to 242, so with this you have practically everything required to compute 3 + 7 + 11 + \ldots + 239.

As a sanity check, you should have (30)(242) = 7260, where you have 30 pairs and each pair sums to 242.

@user1454

“4n-1” was derived from the explanation in the solution to the last question?

This isn’t relevant/required in the solution u posted (as you could ideally tell from my direct response to the solution u posted).

[user1454]

Hello,
I found a part of your response unclear, except for the explanation of the “arithmetic sequence” related to 4n-1. The surrounding explanations were confusing, particularly the term “extrapolate.” Additionally, I didn’t understand why you mentioned the equation y = an + b. Could you provide an example to clarify what you meant by saying “the first difference wasn’t constant, but the second difference was”?

This part of your respone I meant:

@cylverixxx

It’s an arithmetic sequence. If you haven’t learnt that yet then it’s a “sequence of numbers” with the same common difference between its succeeding and preceding terms.

Since the first difference is constant between consecutive terms then you can “extrapolate” a linear function to match the nth term. For example, assume the sequence takes the form y = an + by=an+by = an + b where a,ba,ba,b are coefficients to be found. You know (n,y) = (1,3)(n,y)=(1,3)(n,y) = (1,3) and (n,y) = (2,7)(n,y)=(2,7)(n,y) = (2,7) so you can uniquely find the values of a and b, which happen to be (a,b) = (4,-1)(a,b)=(4,−1)(a,b) = (4,-1). You can use this idea whenever you have a constant first difference (so any kind of sequence like this).

Otherwise, if you’re already familiar with arithmetic sequences then you have: a_n = a_1 + (n-1)dan=a1+(n−1)da_n = a_1 + (n-1)d where (a_1, d) = (3, 4)(a1,d)=(3,4)(a_1, d) = (3, 4). Simplifying should yield a_n = 4n - 1an=4n−1a_n = 4n - 1.

Just as an extra “bonus” lesson for you, if the first difference wasn’t “constant” but the second difference was, then you’d be looking to extrapolate a “quadratic function” for your nth term. The idea generalizes, so basically if nth difference is constant then you can “extrapolate” an nth degree polynomial.

*Additionally, regarding the section where you discussed the number of multiples and pairs, could you explain how this information is useful? I understood your explanation, but I’m curious about how it can help us arrive at the solution to the question.

[cylverixxx]

In an arithmetic sequence, the difference between its succeeding and preceding term is constant.

\begin{array}{|c|c|c|} \hline n & \text{nth term} & \text{First Difference} \\ \hline 1 & 3 & - \\ 2 & 7 & 7 - 3 = 4 \\ 3 & 11 & 11 - 7 = 4 \\ 4 & 15 & 15 - 11 = 4 \\ 5 & 19 & 19 - 15 = 4 \\ 6 & 23 & 23 - 19 = 4 \\ \hline \end{array}

If you wanted to “represent” this with a polynomial, what’s the degree of that polynomial? The most natural choice is a polynomial of degree 1 ( a line) because the first difference is technically just the slope, which is constant for any two consecutive terms.

Now you can imagine finding the equation of a line given the table above. In particular, you want to find (a,b) in y = an + b given the sequence above.

For our specific sequence above (table example above), we know that the slope (first difference) is 4, and using any point (let’s take (2,7)) would yield b = -1. Thus our nth term equation is just y = a_n = 4n - 1

@user1454

Could you provide an example to clarify what you meant by saying “the first difference wasn’t constant, but the second difference was”?

\begin{array}{|c|c|c|c|} \hline n & \text{nth term} & \text{First Difference} & \text{Second Difference} \\ \hline 1 & 4 & - & - \\ 2 & 9 & 9 - 4 = 5 & - \\ 3 & 16 & 16 - 9 = 7 & 7 - 5 = 2 \\ 4 & 25 & 25 - 16 = 9 & 9 - 7 = 2 \\ 5 & 36 & 36 - 25 = 11 & 11 - 9 = 2 \\ 6 & 49 & 49 - 36 = 13 & 13 - 11 = 2 \\ \hline \end{array}

It’s the second difference that’s constant here, so you can’t represent such a sequence with a linear function alone. You’d look towards a quadratic function to “represent” this sequence.

Anyway, if you still don’t get it then I wouldn’t worry too much about it.

[cylverixxx]

@user1454

Additionally, regarding the section where you discussed the number of multiples and pairs, could you explain how this information is useful? I understood your explanation, but I’m curious about how it can help us arrive at the solution to the question.

Wdym, that’s a different solution. You posted the prepswift solution, so I explained how you could use that to show why 3 + 7 + 11 + \ldots + 239 = 7260.

There are 30 pairs which all add up to 242, so we have (242)(30) = 7260.

We surmised that there were 30 pairs because the sequence (3,7,11, \ldots, 239) has the same number of pairs as (4,8,12, \ldots, 240), which we can easily count to have 60 terms and hence 30 pairs. It’s trivial to show that each pair sums to 242.

[user1454]

Hello,

I apologize for my absence recently due to some personal issues. I’ve reviewed everything again, and I believe it all looks good now. Thank you for your thorough explanations!

Just one other question, could you please explain how we calculate the formula based on the second difference?

Also, could you please just tell me where I can find videos regarding “arithmetic and geometric sequences and series” on PrepSwift?

[cylverixxx]

first difference constant → linear function
second difference constant → quadratic function
…
and the trend/pattern continues

Basically, you take three points in the sequence and then you can compute the coefficients of a quadratic that passes through all 3 points.

If you take this example (we had this one above as an example too):

you’re just looking to find a quadratic that passes through (1,4), (2,9), and (3,16). We take exactly 3 points because 3 points uniquely define a quadratic. As you might know, generally n + 1 points are required to uniquely define a polynomial of degree n.

@user1454

Also, could you please just tell me where I can find videos regarding “arithmetic and geometric sequences and series” on PrepSwift?

I’m not sure about this

[Leaderboard]

@user1454

Also, could you please just tell me where I can find videos regarding “arithmetic and geometric sequences and series” on PrepSwift?

The very first video “Sequences I” explains the types of sequences, including arithmetic and geometric sequences.

Application of this (i.e, in problems) are handled depending on the type:

• for arithmetic sequences, they work the same way as the problems on the sum of multiples in an interval, so we expect you to be able to apply those concepts to solve problems involving arithmetic sequences/series.
• for geometric (and other type of) sequences, this is not the case, so either (i) the problem is such that you can do it using pattern recognition or similar deduction, or (ii) sufficient context is provided so that you can apply what’s provided in the question to solve it yourself.

[user1454]

Hello,
Thank you.

By the way, I calculated the values as follows: a = 12/7 , b = 1/7 , and c = 15/7 for this example. I hope this is correct for the equation y = ax^2 + bx + c .

[user1454]

Hello,

Thank you very much! However, I find it a bit strange that, despite watching all three videos on sequences and series, I still find the homework more challenging than what was covered in the lessons. Specifically regarding these two concepts, I’ve learned a lot from asking questions and receiving responses from you and some of your colleagues.

[cylverixxx]

That’s not right. You want to find a quadratic passing through: (1,4), (2,9), and (3,16).

[user1454]

Hi,

Why isn’t it correct? A quadratic equation looks like this: ax^2 + bx + c .

To find the values of a, b, and c, we need to substitute the three points into this equation and solve the resulting three equations.

Could you kindly clarify it explicitly if I happen to be mistaken again?

[user3995]

@user1454

Hello, could you please explain the solution for all these 3 questions below thoroughly?

In the first question, I approached the problem as follows:

1. We have a sequence: n , n+2 , n+4
2. This can be expresed as 600n + (2+4+ …)
3. I set n=3, which gives us : 1800 + (300*1202)
4. However, the answer is 361,200 , which suggests that n is considered to be 1 !
   But since the sequence starts from 3, n should indeed be 3!

I did it this way, not sure if this is also an accepted approach
I found that the 1st term = n
1st term + 2nd term = 2n+2
1st term + 2nd term+ 3rd term = 3n+6
1st term + 2nd term+ 3rd term + 4th term = 4n+12
the coefficient of ‘n’ is always the position we are adding to the sum and the constant is the current position \times previous position so for the addition upto the 600th term it would be 600n+(600*599) = 361,200.

[cylverixxx]

The idea is indeed right, but your execution seemingly was flawed(?)

As you can see, your quadratic only passes through (1,4) and not any of the other points.

[user1454]

Hello,

Yes, thanks I understood, and I think the right answers are : a = 1 , b = -2 , c = 5. I hope so:)