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C1 series question

How would I go about solving this question? Thanks
(edited 7 years ago)
Original post by veebakkas
How would I go about solving this question? Thanks


Hmmm.. which exam board is this? Usually the sum notation is written slightly differently in my experience.
Original post by richpanda
Hmmm.. which exam board is this? Usually the sum notation is written slightly differently in my experience.


I'm using an Edexcel textbook, pretty sure it's outdated though
Original post by veebakkas
How would I go about solving this question? Thanks


Using what you know about summations (particularly the 4 to 30 and 8 to 30 there) which terms cancel out in the equation?
What you're doing is finding the difference between the two Sigmas. Both of them go through values of x from 8 -> 30, but the first one also goes through values of x from 4 -> 7.
Original post by SeanFM
Using what you know about summations (particularly the 4 to 30 and 8 to 30 there) which terms cancel out in the equation?


ngl not sure, the 8 to 30? because that would overlap with the summation of 4 to 30?
Original post by veebakkas
ngl not sure, the 8 to 30? because that would overlap with the summation of 4 to 30?


Precisely :h: you do know it. So if you remove the 8 to 30 from the 4 to 30, what does that leave you with?
Original post by SeanFM
Precisely :h: you do know it. So if you remove the 8 to 30 from the 4 to 30, what does that leave you with?


It's 40, but is there a quicker method? I had to use my calculator for that (I mean algebraically btw)
Original post by veebakkas
It's 40, but is there a quicker method? I had to use my calculator for that (I mean algebraically btw)


Not allowed calculators in C1 :tongue:

The answer to my previous question was that if you take away 8 to 30 from 4 to 30, that leaves you with 4 to 7.

You can then compute that manually (i.e i=47(2x1)=((2×4)1)+((2 times5)1).... \sum_{i=4}^7 (2x-1) = ((2 \times 4) - 1) + ((2\ times 5) - 1).... or use that
i=47(2x1)=[br]i=17(2x1)i=13(2x1) \sum_{i=4}^7 (2x-1) = [br]\sum_{i=1}^7 (2x-1) - \sum_{i=1}^3 (2x-1) using the converse of the logic at the start of the question. You can then use a formula that you know to compute both of those terms.

Or another option if you prefer, split up the sum: i=47(2x1)=2×i=47xi=471=2(4+5+6+7)1111 \sum_{i=4}^7 (2x-1) = 2 \times \sum_{i=4}^7 x - \sum_{i=4}^7 1 = 2* (4 + 5 + 6 +7) - 1 -1 -1 -1
(edited 7 years ago)
Original post by SeanFM
Not allowed calculators in C1 :tongue:

The answer to my previous question was that if you take away 8 to 30 from 4 to 30, that leaves you with 4 to 7.


Why is it 4 to 7, instead of 4 to 8? am I missing something?
Original post by veebakkas
Why is it 4 to 7, instead of 4 to 8? am I missing something?


From the definition of sum :smile: remember the bit on the bottom (8) indicates that the first term of the 8 to 30 is x=8.
(edited 7 years ago)
Original post by SeanFM
From the definition of sum :smile: remember the bit on the bottom (8) indicates that the first term of the 8 to 30 is x=8.


oh yeaa i see

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