Question

How do I attempt this problem? I thought if I were to put 2 directly in, the answer should be straight forward (see attached)

Answer 1

You have to plug in 2. Then you add whatever you get to 1. That is the partial sum.

Answer 2

so i have -5^3

Answer 3

divided by 2! ?

Answer 4

Its asking for partial sums not "terms" in the series. All over 2. So you have -125/2+25
=(-125+50)/2=-75/2

Answer 5

(-5^3)/2

Answer 6

where does the extra +25 come from?

Answer 7

Because thats the first term. The partial sum is when you add the two. The first term gave you a 25. So you have to add that to the second term.

Answer 8

For the 3rd partial sum you'll add the first 3 terms, then the first 4 for the fourth partial sum, etc etc.

Answer 9

malevolence has it - the sequence of partial sums is the value of the entire expression up to some n. If we call the function inside the summation f(x), s1 will be f(1), s2 will be f(1) + f(2) = s1 + f(2), s3 will be f(1) + f(2) + f(3) = s2 + f(3)...

Answer 10

So I don't need to evaluate the ! in the demonimator, or does the +25 compensate for that?

Answer 11

2! is only 2. 3! is 6. etc. You need to evaluate the factorials. The 25 has NOTHING to do with the factorial. It has to do with the fact that it is the FIRST term and you want the SECOND partial sum.

Answer 12

For the four factor, i have (-5^4)/2 + (-37.5) +25. is this the correct way of doing it?

Answer 13

oh, I mean (-5^4)/6

Answer 14

Close, but not exactly. The third term of your series will be s3 = (-5)^4 / 3! + (-5)^3 / 2! + (-5)^2/1!

You can either solve this directly: 625/6 + -125/3 + 25

Or you can remember that your second term was (-5)^3/2! + (-5)^2/1! = -37.5 and substitute this in to the above expression to get:

s3 = (-5)^4 / 3! - 37.5 = 625/6 - 37.5

Answer 15

So the beginning 25 doesn't tag along in the next set. I only use, say, S1 and S2 for the answer to S3.

Answer 16

Yes, each sn is going to be the function inside the sum added to s(n-1)

Answer 17

Thanks everyone!

Answer 18

one more question if you are still there: why isn't this working?
(-5)^5 / 4! + (-5)^4 / 3!

Answer 19

Sorry, I was answering another problem. Remember, s4 is going to be f(4) + s3, NOT f(4) + f(3). So s4 will be
(-5)^5/4! + (-5)^4/3! + (-5)^3/2! + (-5)^2/1!
or
(-5)^5/4! + (625/6-37.5)

Answer 20

(-5)^5/4! + (S3 -S2)

Answer 21

No, (-5)^4/3! is not the value of s3. s3 is the sum of all the terms up to n = 3, not just the term AT n=3

Answer 22

oh, okay

Answer 23

ilovemath7, if you went to do s10000 you would have to do s1+s2+s3....+s9998+s9999+s10000
So however many "s"'s you go out, you have that many terms to add.