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    Hey.
    Please can someone explain how i can use Sn = n/2( 2a + (n-1)d) to find the sum of consecutive numbers, consecutive even/odd numbers, or point me to where i can find out.
    I don`t quite understand how to manipulate the equation.
    Thanks in advanced.
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    (Original post by SamuelN98)
    Hey.
    Please can someone explain how i can use Sn = n/2( 2a + (n-1)d) to find the sum of consecutive numbers, consecutive even/odd numbers, or point me to where i can find out.
    I don`t quite understand how to manipulate the equation.
    Thanks in advanced.
    how many numbers are we talking?
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    (Original post by SamuelN98)
    Hey.
    Please can someone explain how i can use Sn = n/2( 2a + (n-1)d) to find the sum of consecutive numbers, consecutive even/odd numbers, or point me to where i can find out.
    I don`t quite understand how to manipulate the equation.
    Thanks in advanced.
    In each case, work out what n is, work out what a is and work out what d is.

    Eg keeping it simple, how do you find the sum of 1,2,3? How about 1,3,5? 2,4,6?
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    (Original post by SeanFM)
    In each case, work out what n is, work out what a is and work out what d is.

    Eg keeping it simple, how do you find the sum of 1,2,3? How about 1,3,5? 2,4,6?
    Thanks. i`m still finding it difficult still to apply it to summation of larger series

    S3=(1/2)(3)(2+(3-1)1)

    S3=(1/2)(3)(2+(3-1)2)

    S3=(1/2)(3)(2(2)+(3-1)2)

    Do i just derive the 3 formula above, then sub in a different N value?
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    (Original post by SamuelN98)
    Thanks. i`m still finding it difficult still to apply it to summation of larger series

    S3=(1/2)(3)(2+(3-1)1)

    S3=(1/2)(3)(2+(3-1)2)

    S3=(1/2)(3)(2(2)+(3-1)2)

    Do i just derive the 3 formula above, then sub in a different N value?
    Yes, those 3 are correct. An example of something that you're struggling and how you've tried to tackle it would help

    I'm not sure what you mean by that last bit.
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    (Original post by SeanFM)
    Yes, those 3 are correct. An example of something that you're struggling and how you've tried to tackle it would help

    I'm not sure what you mean by that last bit.
    Sorry, i just meant if a question asked for the sum of the the even numbers from 0 to 100 etc, what value of N would i use.

    Ill try and find an example.
    Thanks.
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    (Original post by SamuelN98)
    Sorry, i just meant if a question asked for the sum of the the even numbers from 0 to 100 etc, what value of N would i use.

    Ill try and find an example.
    Thanks.
    Oh yes, I know what you mean - for example - the sum of all even numbers from 2 to 100 inclusive.

    Right, a and d are straightforward - they are both 2.

    But n is slightly less easier. What I would suggest is thinking about it / using your fingers like this:

    '2 is the first number, 4 is the second, 6 is the third, 8 is the fourth, 10 is the fifth... 12 is the sixth, 14 is the seventh, 16 is the eighth, 18 is the ninth and 20 is the 10th. So the one that ends in '10' are multiples of 5, is what we learned from this set of numbers. So using that, 30 is the 15th, 40 is the 20th, 50 is the 25th.. double that and you get that 100 is the 50th term. So n is 50 in this case.
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    As a more formulaic alternative: a = 2 + (n-1)(2) = 2n is the sequence that generates even numbers.

    So the 100 th term is when 100 = 2n \iff n = 50
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    (Original post by SeanFM)
    Oh yes, I know what you mean - for example - the sum of all even numbers from 2 to 100 inclusive.

    Right, a and d are straightforward - they are both 2.

    But n is slightly less easier. What I would suggest is thinking about it / using your fingers like this:

    '2 is the first number, 4 is the second, 6 is the third, 8 is the fourth, 10 is the fifth... 12 is the sixth, 14 is the seventh, 16 is the eighth, 18 is the ninth and 20 is the 10th. So the one that ends in '10' are multiples of 5, is what we learned from this set of numbers. So using that, 30 is the 15th, 40 is the 20th, 50 is the 25th.. double that and you get that 100 is the 50th term. So n is 50 in this case.
    Cheers, very helpful.
    Would i use a similar idea for Odd numbers?
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    (Original post by SamuelN98)
    Cheers, very helpful.
    Would i use a similar idea for Odd numbers?
    Challenge for you then, what is the sum of the odd numbers from 2 to 50?
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    (Original post by SeanFM)
    Challenge for you then, what is the sum of the odd numbers from 2 to 50?
    sn=(n/2)(2a+(n-1)d)

    S23= (23/2)((2(3)+(22)(2))=690 ???
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    (Original post by SamuelN98)
    sn=(n/2)(2a+(n-1)d)

    S23= (23/2)((2(3)+(22)(2))=690 ???
    The key thing is getting a, n and d right I am sure you can put it in the formula correctly. (and of course, finding n is the hardest bit).

    Again, 3 is the first term, 5 is the second, 7 is the third, 9 is the fourth, 11 is the fifth, 13 is the sixth... 23 is the eleventh term, 43 is the twenty first term, 45 is twenty two, 47 is twenty three and 49 is twenty four. So n = 24 in this case.
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    (Original post by SeanFM)
    The key thing is getting a, n and d right I am sure you can put it in the formula correctly. (and of course, finding n is the hardest bit).

    Again, 3 is the first term, 5 is the second, 7 is the third, 9 is the fourth, 11 is the fifth, 13 is the sixth... 23 is the eleventh term, 43 is the twenty first term, 45 is twenty two, 47 is twenty three and 49 is twenty four. So n = 24 in this case.
    Ah ok. Thanks.
 
 
 
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