4 Sequences: Floor(sqrt(n)); a=1,b=5,c=a+b,etc; 1,1+1,(1+1)*1,(1+1)*1+2,etc; ...

[Setzana]

I'm stuck on some of these sequences. I know that the little L shaped braces means 'floor value', but I'm not completely sure that I understand what that means. I think it just means to round down, right? But to what place value? Whole number I'm guessing? This technically isn't a math class, it's math for computer science, so the book already expects us to know some of this, but I don't. Also, I'm non-traditional and it's been a few years since I've had algebra. I'm struggling so bad with these sequence problems, my brain just doesn't work this way. I've got the first problems already, but these are where I'm getting stuck.

List the first 10 terms of each of these sequences:

d. The sequence whose nth term is ⌊√n⌋

So on this one, would I just start by sticking numbers starting at 1 in for n, find the square root of those numbers, and then if it's a decimal round down to the nearest....whole number, maybe?

e. The sequence whose first two terms are 1 and 5 and each succeeding term is the sum of the two previous terms

The wording of this is a little confusing.

Lets see if I'm understanding this.... 1, 5 and then the next would be 1+5 = 6 and the next is 5 + 6 = 11 6+11 =17 is that right?

g. The sequence whose terms are constructed sequentially as follows: start with 1, then add 1, then multiply by 1, then add 2, then multiply by 2, and so on.

So start with 1 + 1 * 1 + 2 * 2 what? How does this result in anything more than an unending problem? Each of those steps is separated by a comma, not a period which would signal an end to each set of steps. That doesn't make sense.

h. The sequence whose nth term is the largest integer k such that k! ≤ n

I don't understand what this problem is trying to say, or what it wants. Zero comprehension of this one.

 

[Setzana]

Find the sequence

#6 Find the first 10 terms of each of these sequences:

d. The sequence whose nth term is ⌊√n⌋
g. The sequence whose term sare constructed sequentially as follows: start with 1, then add 1, then multiply by 1, then add 2, then multiply by 2, and so on
h. The sequence whose nth term is the largest integer k such that k! ≤ n

On letter d I'm just not sure where to start because if I plug 1 in for n the sq rt of 1 is 1, and for two it rounds down to 1, and same for 3 and 4 is 2 etc. so the first four numbers of the sequence is 1, 1, 1, 2 ....it just doesn't seem right. I don't know. I hate sequences, this stuff doesn't translate well in my head.

On letter g, that's exactly how the problem is written, punctuation and all. It confuses me because it uses commas so I'm not sure how to make that into anything other than just one long unending problem. As it's written, it would be 1 + 1 * 1 + 2 * 2 I'm not understanding how I'm supposed to come up with 10 terms, or where the problem is "supposed" to break apart itno different mini problems to come up with 10 numbers. Also the pattern changes. It's 1 + 1 * 1 and so shouldn't the next part be 2 + 2 * 2?

For letter h, I honestly have no clue what it's wanting. I'm not understanding the problem, or what the solution should look like, or any of it.

Any help is much appreciated.

 

[ksdhart2]

For problems d and e, you have the correct answer but you're overthinking it and psyching yourself out. For problem g, I agree it is poorly worded. I'd ask your professor to double check, but I see it as saying the sequence is defined as follows:

f(1) = 1
f(2) = f(1) + 1
f(3) = f(2) * 1
f(4) = f(3) + 2
...
f(10) = ???

And finally, on problem h, the best place to start is: do you know the definition of k! (read as "k factorial")? If not, you might find this page helpful.

 

[stapel]

Can't post this reply

List the first 10 terms of each of these sequences:

h. The sequence whose nth term is the largest integer k such that k! ≤ n

I don't understand what this problem is trying to say, or what it wants. Zero comprehension of this one.

Follow the definitions, and see there they lead. You know that k! = 1*2*3*...*k, and you know that you're supposed to find k such that k! is less than n for each term a_n, so:

 

[stapel]

first term: n = 1:

3! = 1*2*3 = 6 > 1 = n (too big)
2! = 1*2 = 2 > 1 = n (too big)
1! = 1 = 1 = n

 

[stapel]

second term: n = 2:
. . .3! = 6 > 2 = n (too big)
. . .2! = 2 = 2 = n

Then the second term must be a₂ = k = 2.

third term: n = 3:
. . .3! = 6 > 3 = n (too big)
. . .2! = 2 < 3 = n

Then the third term must be a₃ = k = 2.

And so forth. :wink: