C1 series question

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    How would I go about solving this question? Thanks
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    (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.
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    (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    I'm using an Edexcel textbook, pretty sure it's outdated though
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    (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?
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    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.
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    (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?
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    (Original post by veebakkas)
    ngl not sure, the 8 to 30? because that would overlap with the summation of 4 to 30?
    Precisely you do know it. So if you remove the 8 to 30 from the 4 to 30, what does that leave you with?
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    (Original post by SeanFM)
    Precisely 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)
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    (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

    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  \sum_{i=4}^7 (2x-1) = ((2 \times 4) - 1) + ((2\ times 5) - 1).... or use that
     \sum_{i=4}^7 (2x-1) = 

\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:  \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
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    (Original post by SeanFM)
    Not allowed calculators in C1

    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?
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    (Original post by veebakkas)
    Why is it 4 to 7, instead of 4 to 8? am I missing something?
    From the definition of sum remember the bit on the bottom (8) indicates that the first term of the 8 to 30 is x=8.
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    (Original post by SeanFM)
    From the definition of sum 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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