#summation notation
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You have to write like this (2(1)+3) +(2(5)+3)+ (2(7)+3)+ (2(9)+3)=??
then what number would i stop at
Simplify the number @pallid summit
what do you mean?
I mean simplify the expression
Whatever, The number would be 56.
What grade are you study?
= 2(1)+3+2(2)+3+2(3)+3+...2(n)+3
=2(1+2+3+...+n)+3(n)
if you know AP
then,
sum of first n natural numbers is n(n+1)
so,
1+2+3+...+n=n(n+1)/2
so our expression equals to
=2(n(n+1)/2)+3n
=n(n+1)+3n
=n^2+n+3n
so answer is
$n^2+4n$
Pro_Hecker
and btw by write out each sum they meant to solve sum for questions 17-20
type .close to close this thread if you're problem has been resolved
so then i would just write n^2 + 4n?
n is an ambiguous number
As you keep adding things up, n will be the last number you use in the summation
The people here are giving examples of numbers you can use to represent n. But if you go all the way to n sums, there's a pattern that will appear.
But to answer your original question:
You don't use commas for this. Since you're doing a sum, use + instead
i gave answer its n^2 +4n
Sure, but I don't think the OP fully understand why so I'm explaining the concept to her
then how would you write the sum for
would the concept be the same as the previous one?
Yeah, except here you begin at 2 instead of 1
how would you include ln k into the sum?
You can imagine a big parenthesis that covers everything to the right of the sum symbol
So you can start with $$(-1)^2 \ln{2}$$
plus the next number + the number after that
Mellow
but n is the variable on top so the calculations would continue to go on..?
Yeah it would continue to go until it reaches n
so then how would i calculate the sum?
There's a chance that you can't
You might have to leave it terms of n
For stuff that has variables like this, there's going to be a pattern that the sum follows. If you find that pattern, you can turn it into an equation
It can get closer and closer to a single value
We say it "converges" if it does
There are times where you can't tho. We say it "diverges".
At that point, you have to write the pattern with some ... in the middle to say this goes on for a while
we havent gotten to the divergent convergent part so i think the teacher wants us to write an equation representing the sum
but im not sure how or where to start
Ahh okay, I think Ik where you're at then
Well you start with $$(-1)^2 \ln{2} + ... $$ what comes after?
Mellow
(-1)^3 ln3
(-1)^4 ln4
Right you got it
So what you have rn is $$(-1)^2 \ln(2) + (-1)^3 \ln(3) + (-1)^4 \ln(4) + ... $$
Mellow
If you keep doing this until n, what does it look like?
Aka what do you get for the last one?
Yes, just waiting on the OP's response
(-1)^k ln(k)??
can you DM me the answer
Close! But remember that k becomes the number at the top at the end
so replace k with n?
Sure, but it's simpler than you think
(-1)^n ln(n)
Exactly!
So then wouldn't the summation look like $$(-1)^2 \ln(2) + (-1)^3 \ln(3) + (-1)^4 \ln(4) + ... + (-1)^n \ln(n)$$
Mellow
It can't be like the first one cause the number only keeps on getting bigger
Pro_Hecker
but the first one is similar since the top number/variable is n
there is no solution
Yeah that's true but that's because the first one had a pattern to it
Post marked as solved by @olive solar.
Use .unsolved if this was a mistake.