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    Find the sum of the first 50 even numbers. (sequences)

    I used the sum of natural numbers formula like this:

    Sequence: 2, 4, 6......50

    so: 2 ( 1, 2 , 3.....25)

    {n(n+1)}/2 is 2[ {25(26)}/2 ] which is 650.

    The answer is 2450.... :confused:
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    (Original post by littlemisshala)
    Find the sum of the first 50 even numbers. (sequences)

    I used the sum of natural numbers formula like this:

    Sequence: 2, 4, 6......50

    so: 2 ( 1, 2 , 3.....25)

    {n(n+1)}/2 is 2[ {25(26)}/2 ] which is 650.

    The answer is 2450.... :confused:
    whoops

    its from n = 1 to 50; you were thinking sum of even numbers from 1 to 50...
    and so was I
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    (Original post by kikzen)
    the answer is 650 (manually work it out if you like)... check what youre reading?
    I worked it out to be 650 but the book says its 2450!!! Definite error?
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    (Original post by littlemisshala)
    Find the sum of the first 50 even numbers. (sequences)

    I used the sum of natural numbers formula like this:

    Sequence: 2, 4, 6......50

    so: 2 ( 1, 2 , 3.....25)

    {n(n+1)}/2 is 2[ {25(26)}/2 ] which is 650.

    The answer is 2450.... :confused:
    even numbers can be worked out using 2n, so the last term is 100

    using Sum_n=(n/2)(a+l) for arithmetic series)

    a = first term = 2(1) = 2
    l = last term = 2(50) = 100
    n = 50

    Sum_50 = (50/2)(2+100)
    Sum_50 = 25*102
    Sum_50 = 2550

    The answer in the book suggests the first even number is 0, in which case...

    a = first term = 0 = 0
    l = last term = 2(49) = 98
    n = 50

    Sum_50 = (50/2)(0+98)
    Sum_50 = 25*98
    Sum_50 = 2450
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    (Original post by El Stevo)
    even numbers can be worked out using 2n, so the last term is 100

    using Sum_n=(n/2)(a+l) for arithmetic series)

    a = first term = 2(1) = 2
    l = last term = 2(50) = 100
    n = 50

    Sum_50 = (50/2)(2+100)
    Sum_50 = 25*102
    Sum_50 = 2550

    The answer in the book suggests the first even number is 0, in which case...

    a = first term = 0 = 0
    l = last term = 2(49) = 98
    n = 50

    Sum_50 = (50/2)(0+98)
    Sum_50 = 25*98
    Sum_50 = 2450
    I don't understand what you mean by 2n? Also, why do u use the above formula and not the sum of natural no.formula?
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    (Original post by littlemisshala)
    I don't understand what you mean by 2n? Also, why do u use the above formula and not the sum of natural no.formula?
    He used 2n because it is even numbers, ie numbers which have 2 as a factor.
    So when;
    n = 0, 2n = 0
    n = 1, 2n = 2
    n = 2, 2n = 4
    n = 3, 2n = 6

    e.g. even numbers.

    if you wanted numbers which have 3 as a factor, you would use 3n.

    If you wanted, you could derive a formula for the sum of even numbers.

    Sn = n/2(2(2) + 2(n-1))
    Sn = n/2(4 + 2n - 2)
    Sn = n/2(2n + 2)
    Sn = n^2 + n

    E.g.
    S4 = (4)^2 + 4 = 20

    2 + 4 + 6 + 8 = 20

    S7 = (7)^2 + 7 = 56

    2 + 4 + 6 + 8 + 10 + 12 + 14 = 56
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    (Original post by littlemisshala)
    Find the sum of the first 50 even numbers. (sequences)

    I used the sum of natural numbers formula like this:

    Sequence: 2, 4, 6......50

    so: 2 ( 1, 2 , 3.....25)

    {n(n+1)}/2 is 2[ {25(26)}/2 ] which is 650.

    The answer is 2450.... :confused:
    El stevo is right.
    the question ask SUM OF FIRST 50 EVEN NUMBER,
    50 IS THE 26TH EVEN UUMBER,98 IS THE 50TH
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    Sequence 0,2,4,6,8 etc..

    Sn=n x (a+l)/2

    a=2
    l=98
    n=50

    S50=50 x (98+0)/2
    S50=50 x 49
    S50=2450.
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    use the formula : Sn = n/2 (2a + (n-1)d))
    where
    n=50
    a=0
    d=2

    s50= 50/2 (2x0 +(50-1)2))

    S50= 25 x 98 = 2450
 
 
 
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