Limit of sequence

[plush oasis]

I have no idea how the book changes the numerator into the sum expression.
It would be great if anyone can show the steps!

[plush oasis]

The question is to solve the limit when n approaches to infinity, therefore I classified it as Calculus.

[unborn drift]

Do you know how to do summation/Sigma?

[plush oasis]

Yes

But I couldn’t understand how it is written like this🤔

[unborn drift]

Same here my guess is probably the pattern that they give.

Since you know the Sigma notation of the bottom one, right?

[plush oasis]

Yes

😉

There’s also a note besides a solution.

Let me take a picture of it

The “分子” means the numerator

When I expand the numerator part of the original question, there’s already a 1^2+2^2+…+(2n)^2. Apparently, it is correct. But how come I need to subtract 1^2+2^2+…+n^2 ?

[unborn drift]

I think it is to make the n subject as 2n^2- n^2 = n^2. The real value will be zero. Idk ask the <@&286206848099549185> for deeper understanding

Sorry

[plush oasis]

It’s ok! I’ll keep trying. Thank you so much😊

[unborn drift]

No problem.

[plush oasis]

I solved it in another way, but I am still confused by how it changes into that form.
Maybe I’ll ask my mathematics teacher after class. 🧐

[unborn drift]

Yeah do that.

[plush oasis]

[brittle steeple]

summing from n+1 to 2n is the same as summing from 1 to 2n and subtracting the sum from 1 to n

feel free to ping if its still unclear

[little lichen]

u do that to start ur sum in numerator from (n+1)^2

[cursive eagle]

Your solution is completely fine, i think the idea is to make computation a little bit lighter because in the picture you directly have the formula for every bit and also its equal to "the left bit-1"

@plush oasis

[plush oasis]

Thanks for all your help!
I have a clearer understanding of the solution. I appreciate that.

.solved