Find the 9th partial sum of the summation of 3i-4, from i equals 1 to infinity. I think I know how to do it but I can't tell if it's arithmetic or geometric so I know which formula to use.
You could try breaking this into 2 separate summations. Let the first one be the sum (i=1 to i=9) of 3i. This would be the same as "three times the sum from i=1 to i=9 of i alone." The other sum would be simple to find: 9 times 4. Don't forget to put a - sign in front of this sum.
What is the sum of \[\sum_{1}^{9}i~?\]
There's a formula for that. although I don't recall it at the moment. what is the sum of i when i goes from 1 to 1? It's just 1. what is the sum when i goes from 1 to 2? It's 3. Go ahead and experiment. Have fun. Show all work.
Would it just be 9 for that part then? I'm not quite sure how to do it with the i
I ranges from 1 to 9, and we're adding them up. What do you get? The 2nd partial sum would be 1+2=3; the 3rd would be 1+2+3=6, and so on.
OKay, so then it would be 1+2+3+4+5+6+7+8+9 which would be 45 right?
To me, if you jot down some numbers , you can see what is going on Like this: if i =1, then the first number is 3*1 -4 =-1 if i =2, then the second number is 3*2 -4 = 2 do some more, you can see it is a arithmetic series, and it is easy
Okay, so then because it's an arithmetic series, we have to use \[\frac{ n }{ 2 }(a _{1}-a_{9})\]
we have another formula for it
Reverse the order. a9 - a1.
Why would it be the other way around?
or you can use it but have to find a9 first
4.5(-1+23) which would be 99
Technically both ways would work I guess; if you had a9 and then a1
Because the i values increase. if y ou were to subtract 9 from 1, you'd get -8. It's unlikely that that negative number would lead you to the correct sum. You're free to use whichever approach you want. I'd recommend you separate the addends and create 2 summations: 3 times the summation from 1 to 9 of i, and then subtract the summation from 1 to 9 of 4.
summation is by definition addition (or combination, if both pos. and neg numbers are involved). That alone suggests that you have an arith. sequence here, not a geom sequence. Geom sequences involve multiplication and exponentiation.
Alright, so then would the answer I got be correct?
yup
Alright, thanks. I'm also kind of having trouble with another question, if you have time to spare at all.
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