Question

how do i do number 1??
answer for one is (3/2n) + (1/6n^2) + (4/3)
(for the first part of the question)

Answer 1

wait, so what have you solved? You want the answer to sum k = 1 to n of (1/n * [k^2/n^2 + 2k/n])?

Answer 2

i don't get how that's the answer..

Answer 3

like i've tried separating them into 3 summations but idk how simplify it

Answer 4

That can't be the answer since we already took n -> infinity....

Answer 5

Oh I'm sorry you want to compute the sum first, then the limit.

Answer 6

yes compute first. i'm pretty confident ik how to take the limit thou. just the first part got me stumped

Answer 7

Having established what the answer is can you use induction?

Answer 8

induct over n

Answer 9

my teacher has not gone over induction. all i have is the summation formulas..maybe if u explain what induction is i can tell u if it's something we've brushed over

Answer 10

ah kk sec

Answer 11

ok btw thanks for the help(:

Answer 12

So, show that the sum = (3/2n) + (1/6n^2) + (4/3) for n = 1. That's easy. Then show that if it's true for n = m (arbitrary m), then it's true for n = m+1. That is, ASSUME sum k = 1 to m (expression in problem, substituting m for n) = (3/2m) + (1/6m^2) + 4/3. Then show given that, you get sum k = 1 to m+1 (expression given in problem, but now substitute m+1 for n) = 3/2(m+1) + 1/6((m+1)^2) + 4/3. If you do that, you know that if the expression holds for n = any number m, it holds for the next one (m+1). Having shown it is true for n = 1, it's true for all n. Note that this does NOT mean that it is true for the limit in general. That must be done separately. Induction holds in the finite case.

Answer 13

Ah there's a trick

Answer 14

ok i understand everything but where did/ how did u derive (3/2n) + (1/6n^2) + (4/3)?

Answer 15

Oh, isn't that what you said?

Answer 16

yea it's just the answer u're supposed to reach from the answer key. I just didn't know how u get to that answer

Answer 17

The idea is that you show both that it holds for n = 1 and you show that assuming it holds for n = m, it holds for n = m+1.

Answer 18

Ah you're probably not supposed to use this technique then. Also it works best when you can easily compare the two sums, which is difficult in this case. I did get sum k = 1 to m+1 (1/(m+1)(k^2)/(m+1)^2) + 2k/(m+1)

Is the same as

sum k = 2 to m+2 (1/(m)((k-1)^2)/(m)^2) + 2(k-1)/(m), but then the k terms are still hard to compare. So nvm with that approach

Answer 19

ok. so what should be my first step since i shouldn't use that approach? i'm sorry lol it's super late and im tried ,brains not with me atm

Answer 20

ooooooooooooh

Answer 21

You know formulas for sum i = 1 to n of i, and sum i = 1 to n of i^2 right?

Answer 22

yes we are on the same page now lol

Answer 23

Yea, they're in terms of n

Answer 24

do u know those cuz i have them handy if u need them

Answer 25

Basically all you have to do is divide by an additional n^2, or n^3, depending on the term

Answer 26

Since you have here sum k = 1 to n of k^2 / n^3, and 2k/n^2

Answer 27

So take the formula for the k^2, and then divide by n^3. Take the 2k/n^2, and multiply by 2/n^2

Answer 28

Since n is constant, you can do this.

Answer 29

(It wouldn't be ok to do this with a variable like k)

Answer 30

wait why would it be 2k/n^2 for the last term?

Answer 31

Sorry, I mean take the formula for sum k = 1 to n of k, and then multiply by 2/n^2

Answer 32

So in other words 1/2 n (n+1) / n^2

Answer 33

Oh sorry, times 2

Answer 34

You do likewise with the other term (the k^2 one) just make sure you multiply by 1/n^3 instead of 2/n^2

Answer 35

haha ok give a quick second to try and process

Answer 36

This is a nice problem. Sorry I went a bit crazy when I got it hehe

Answer 37

I got ((n+1) / n )+((n+1)(2n+1)/(6n^2)), which is equal to the answer you gave originally

Answer 38

haha nah it's all good. i think i'm sort of getting it now. thanks!

Answer 39

Those terms are what I get for sum k = 1 to n of k/n^2, and sum k = 1 to n of 2k/n^3, respectively. Again, use the fact that sum k = 1 to n of k/n^2 = n^2 sum k = 1 to n of k, and likewise for the other one.

Answer 40

got it. Thank you so much!! really appreciate your time(:

Answer 41

so my teacher realized that he didn't teach this to us yet lol...but thanks to you I comprehended it so easily @elementalmagicks