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FP1: series

why in the series equation is there r = 1 at the bottom but an n at the top? what do either mean? and why cant I use it in the form of r to calculate b, but have to use part a (in terms of n)? even though there is an r at the bottom?
Screen Shot 2018-03-16 at 16.02.02.png106.0KB
Original post by Maths&physics
why in the series equation is there r = 1 at the bottom but an n at the top? what do either mean? and why cant I use it in the form of r to calculate b, but have to use part a (in terms of n)? even though there is an r at the bottom?


This is covered in C2... the bottom limit represents the first term of the summation, and the upper limit represents the last term in the summation.

Not even sure what you're asking after that. We have ∑r=1120=∑r=120−∑r=110\displaystyle \sum_{r=11}^{20} = \sum_{r=1}^{20} - \sum_{r=1}^{10}
Original post by RDKGames
This is covered in C2... the bottom limit represents the first term of the summation, and the upper limit represents the last term in the summation.

Not even sure what you're asking after that. We have ∑r=1120=∑r=120−∑r=110\displaystyle \sum_{r=11}^{20} = \sum_{r=1}^{20} - \sum_{r=1}^{10}


its been a while since ive done c2....cool.

when its in the form r: 6r2+4r−16r^2 + 4r -1, why do we have to have in the form n - whats the difference?
Original post by Maths&physics
its been a while since ive done c2....cool.

when its in the form r: 6r2+4r−16r^2 + 4r -1, why do we have to have in the form n - whats the difference?


6r2+4r−16r^2+4r-1 represents the rth term of the sequence ur=6r2+4r−1u_r = 6r^2+4r-1

∑r=1n6r2+4r−1\displaystyle \sum_{r=1}^n 6r^2+4r-1 is a series, which means summing up the terms of the sequence uru_r starting from r=1r=1 up to r=nr=n.
Original post by RDKGames
6r2+4r−16r^2+4r-1 represents the rth term of the sequence ur=6r2+4r−1u_r = 6r^2+4r-1

∑r=1n6r2+4r−1\displaystyle \sum_{r=1}^n 6r^2+4r-1 is a series, which means summing up the terms of the sequence uru_r starting from r=1r=1 up to r=nr=n.


youre the best! thanks

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