from what i'm gathering i would just use the rule of sums

>>8592517
you goofed in that you replaced delta x with 2+3/n.
x_i also isnt 2+3/n but 2+3i/n.

>>8592542
delta x is 3/n.
x_i is 2 + 3 i /n
plug those into the sum, get sum sums of square... you just need the formula for sum i^2 and sum i.

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>>8592542
ah my bad. ok i think this works better.

>>8592553
you need a pair of parentheses.

\sum_{k=1}^n (4-2*(2+3 k / n ))*3/n = -9-9/n.
As n goes to infinity, the sum goes to -9.

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>>8592566
like this?

sorry i didn't mean to try and parrot. it's just that i can't see the simplification approach. i still end up with n and i in my expression. ill show what i got.

>>8592576
ah sorry the summation doesn't apply there since there's no i. damn i keep making mistakes.

you should really get your arithmetic in check before attempting to do analysis.
the last line is a complete clusterfuck

>>8592581
at the second to last line you should multiply everything out and then to evaluate the summation you just substitute n(n+1)/2 for i. At least I think. It gave me the right answer anyway.

>>8592576
yes, that's it I think. you need to use 1+2+...+n = n*(n+1)/2 and 1+2^2 + ... + n^2 = n*(n+1)*(2n+1)/6 and algebra.

>>8592654
No need for the other summation formula. Just simplify to -18*i/(n^2) and then pull the -18/n^2 out of the summation, then evaluate the sum from 1 to n of i, which is just n(n+1)/2. Simplification from there leads to -9 - 9/n

>>8592642
you mean expand? i thought i was supposed to take 3/n out of the summation since it has no i to it.

>>8592688
You can expand to get the sum from 1 to n of -18*i/n^2 and then you can take -18/n^2 out of summation and you're left with the sum of i from 1 to n. Which is just the summation formula n(n+1)/2. Multiply that with -18/n^2 and then you can evaluate the limit. You can take the 3/n out immediately if you want, it makes no difference.

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>>8592705
alright i did it myself but i seem to have gotten myself still stuck with n over n.

>>8592795
oops nevermind im stupid. it cancels out to being 1+1/n. so the answer then if i evaluate the limit is -9*(1+0)=-9 yes?

sorry for bothering everyone. looks like i still have to practice.

>>8592517
Can I ask what's the use in that ? The integral is simple, why can it in that monstrosity of a sum ?

>>8592812
Change*

>>8592812
to define it in another way

>>8592812
isn't that the definition of a definite integral?

>>8592812
CUZ FUCK U THATS WHY BITCH

>>8592824
Fundamental theorem of Calculus would help immensely here

>>8592863
don't you need riemann sums to prove that?

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i tried to do another one and this time im really boned about factoring out i to sub in sum of squares.

>>8593193
Start by factoring out 1/n^3. This will leave you with (i^3/n - 3i^2) in the summation. Now just split up that summation into two summations and factor out of each of those. Your two summations will be of i^3 and i^2. There are formulas for those sums.

>>8593215
can i just sum of squares and just cube it.

>>8593221
Not sure what you mean

>>8593226
like if i substitute i^3 and i^2 with [n(n+1)/2]^3 and ^2 respectively.

>>8593230
No that won't work. Weirdly if you square the formula n(n+1)/2 you get the formula for i^3.

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>>8593249
i see. does 1/n^3 have to be applied to both summation?

like i have this.

>>8593260
Yes the 1/n^3 must be distributed to both summations, so you will need parentheses that enclose both to distribute that 1/n^3.

>>8593265
k, other than that is it valid?

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>>8593266
Okay I was trying to evaluate what you wrote but it wasn't working. You forgot to put that the i^3 term will have its own 1/n.

So it should be this:

>>8593292
wew lad that looks like a pain to expand.

>>8593298
it's not that bad since you're evaluating a limit and all the lesser powers of n are irrelevant as n goes to infinity.

When I see that I just pay attention to the highest powers of n and forget about the rest, so it becomes 1/4 - 6/6 = -3/4 = -0.75 which agrees with the result of the definite integral.

>>8593308
wouldn't it be 1/4-3/6 since it's 3(n+1)(2n+1)/6?

>>8593372
3n*n*2n = 6n^3