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Mathematics 8 Online
OpenStudy (anonymous):

Infinite sequences and summations Σ6 k=3 (k-5)/(k-1) the answer is -17/15 and Σ8 k=1 (1/2k+7) ans=74 I missed this lesson and I need help on how to get the answer! Please help!

OpenStudy (loser66):

I think there is no one can help you with this stuff because it is toooo non-sense. you missed something from the notation. the sum (sigma) has to have something else than just that. It is infinite sequence, how can it has summation, just limit of partial sum. So, if you really need help, you must give more information

OpenStudy (anonymous):

the 6 is on top of the sigma and the k=3 is below the sigma. I was not sure how to write it that way on here D: and the (k-5/k-1) was next to the sigma. As I said, I missed this lesson so I'm not sure where to start.

OpenStudy (loser66):

ok, I think I can help

OpenStudy (loser66):

\[\sum_{k=3}^{6}\frac{ k-5 }{ k-1 }\]is it yours?

OpenStudy (anonymous):

Yes, that's it!

OpenStudy (loser66):

the \[\sum_{k=3}^{6}\] is the notation indicate to the number of terms you have to calculate, it means you have to calculate the 3rd number + 4th +5th +6th . stop at 6th. got it?

OpenStudy (anonymous):

I'm with you so far.

OpenStudy (loser66):

you just plug the number of k into the function, i mean, when k =3 -----> (3-5)/ (3-1) k =4------> (4-5)/(4-1) k=5---> (5-5)/ (5-1) k=6 -------> (6-5)/(6-1 ---------------------------------------- add them up ------> = -1 + (-1/3) +0 + (1/5) = -17/15 do the same with the second one. got it?

OpenStudy (anonymous):

Oh! I see, I will do the calculation and see if I get the same answer.

OpenStudy (anonymous):

Ok! I did the calculation for my own assurance!

OpenStudy (anonymous):

good for you, I am glad to help people who wants to study

OpenStudy (anonymous):

knock, knock. are you ok?

OpenStudy (anonymous):

\[\sum_{k=1}^{500} 2500\] I understand the last problems. But How would I find the sum of this one, if there is no "k"?

OpenStudy (anonymous):

no k, that means only k1

OpenStudy (anonymous):

the sum of 1 constant is itself

OpenStudy (anonymous):

so, 500*2500? o.O

OpenStudy (anonymous):

k go from 1 to 500, means you will have 500 numbers under the sum, but, if you don't have k to count , your first number is k=1 ---> k-1 =2500, k_2 =0,k_3 =0......so on and so on... so, sum of 2500 + 0+0 +0 +......... = 2500, right? got it?

OpenStudy (anonymous):

sorry, k_1 not k-1

OpenStudy (anonymous):

so then the answer is just 2500?

OpenStudy (anonymous):

yeap

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