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I get sigma notation but how do i do this

It says rewrite following sums in sigma notation
Q) 4,7,10...31

Thanks x
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username2859410
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RDKGames help please
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(Original post by samantham999)
I get sigma notation but how do i do this

It says rewrite following sums in sigma notation
Q) 4,7,10...31

Thanks x
So write out the sequence, it seems to be an arithmetic sequence, so find the formula for the n^{th} term of the sequence in terms of n.

Then find out what term 31 is as that will be your upper limit for the sum, after which point it is pretty straight forward.
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(Original post by RDKGames)
So write out the sequence, it seems to be an arithmetic sequence, so find the formula for the n^{th} term of the sequence in terms of n.

Then find out what term 31 is as that will be your upper limit for the sum, after which point it is pretty straight forward.
Written our sequence: 4,7,10,13,16,19,22,25,28,31
D is 3
A is 4

So does 10 go at the top of sigma and at the bottom what do i write?
There is also brackets that I have to include which is (3r + 1) could you tell me how i get that
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(Original post by samantham999)
Written our sequence: 4,7,10,13,16,19,22,25,28,31
D is 3
A is 4

So does 10 go at the top of sigma and at the bottom what do i write?
There is also brackets that I have to include which is (3r + 1) could you tell me how i get that
You got the formula right, so a_r=3r+1. You get this from saying that a_1=4, a_2=7 so how are these terms related to the result? You get it by inspection. Otherwise, you know the difference is 3, so you start with 3r. Then you plug in r=1 and see how far off your first term you are, then adjust it accordingly; so add 1. Then check if this holds for all the terms.

And yes the 10 goes on the top as 31 is your final term. At the bottom you say which term you begin with, so r=...

Also to note, r and n are interchangeable in this example with sigma. Usually when it comes to sigma the convention is to say the r^{th} term but you can also use n if you so wish.
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(Original post by RDKGames)
You got the formula right, so a_r=3r+1. You get this from saying that a_1=4, a_2=7 so how are these terms related to the result? You get it by inspection. Otherwise, you know the difference is 3, so you start with 3r. Then you plug in r=1 and see how far off your first term you are, then adjust it accordingly; so add 1. Then check if this holds for all the terms.

And yes the 10 goes on the top as 31 is your final term. At the bottom you say which term you begin with, so r=...

Also to note, r and n are interchangeable in this example with sigma. Usually when it comes to sigma the convention is to say the r^{th} term but you can also use n if you so wish.
So 3r is basically the common difference right
And I use trial and error to make the first term?
At the bottom r=1 right? Thanks for this
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(Original post by samantham999)
So 3r is basically the common difference right
Yes


And I use trial and error to make the first term?
It's not so much "trial and error" but it shouldn't be too difficult to spot so you can't really overlook it. Since it's an arithmetic sequence, if you get it for the first one, most of the time other terms will follow easily from it.

At the bottom r=1 right?
Yep, because you are starting your sum from the 1st term.

Thanks for this
No problem
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