Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

Sum the series 1 x 2 + 2 x 3 + 3 x 4 + .... +2019 x 2020?

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  • 6 months ago
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    Let's take each product and call it one term.

    So the first term is 1 * 2 = 2

    The second term is 2 * 3 = 6, etc.

    As we build out the first few terms and see what the sum is up to that point, we get this set of numbers:

    2, 8, 20, 40, 70, 112

    For the sum of the first through 6th term. You're looking for the 2019th term of this sequence.

    So let's see if this can be constructed with a polynomial expression. We'll start testing the differences between the values to see what we get:

    2 - 8 = -6

    8 - 20 = -12

    20 - 40 = -20

    40 - 70 = -30

    70 - 112 = -42

    Now let's get the difference of the differences:

    -6 - (-12) = -6 + 12 = 6

    -12 - (-20) = -12 + 20 = 8

    -20 - (-30) = -20 + 30 = 10

    -30 - (-42) = -30 + 42 = 12

    And one more time, the difference of these differences:

    6 - 8 = -2

    8 - 10 = -2

    10 - 12 = -2

    Now that we have the same value from the third time through this sequence we know that we have a cubic expression (third-degree polynomial). Let's set up that general form:

    f(x) = ax³ + bx² + cx + d

    We have four unknowns so we need to use the first 4 terms (where x = 1 through 4) to create four equations so we can solve for the unknowns. Again the first 4 terms of this sequence is: 2, 8, 20, 40

    Substituting what we know gives us:

    f(1) = a(1)³ + b(1)² + c(1) + d and f(2) = a(2)³ + b(2)² + c(2) + d and f(3) = a(3)³ + b(3)² + c(3) + d and

    f(4) = a(4)³ + b(4)² + c(4) + d

    2 = a + b + c + d and 8 = 8a + 4b + 2c + d and 20 = 27a + 9b + 3c + d and 40 = 64a + 16b + 4c + d

    Solve the first equation for d and substitute into the others:

    2 = a + b + c + d

    2 - a - b - c = d

    8 = 8a + 4b + 2c + 2 - a - b - c and 20 = 27a + 9b + 3c + 2 - a - b - c and 40 = 64a + 16b + 4c + 2 - a - b - c

    8 = 7a + 3b + c + 2 and 20 = 26a + 8b + 2c + 2 and 40 = 63a + 15b + 3c + 2

    6 = 7a + 3b + c and 18 = 26a + 8b + 2c and 38 = 63a + 15b + 3c

    We can simplify the second equation by dividing both sides by 2. Then we'll solve for c and substitute that into the other two:

    18 = 26a + 8b + 2c

    9 = 13a + 4b + c

    9 - 13a - 4b = c

    6 = 7a + 3b + c and 38 = 63a + 15b + 3c

    6 = 7a + 3b + 9 - 13a - 4b and 38 = 63a + 15b + 3(9 - 13a - 4b)

    6 = -6a - b + 9 and 38 = 63a + 15b + 27 - 39a - 12b

    -3 = -6a - b and 11 = 24a + 3b

    Solving the first equation for b and substituting:

    -3 = -6a - b

    -3 + 6a = -b

    3 - 6a = b

    11 = 24a + 3b

    11 = 24a + 3(3 - 6a)

    11 = 24a + 9 - 18a

    2 = 6a

    1/3 = a

    Now we can start working back to solve for b, c, and d:

    b = 3 - 6a

    b = 3 - 6(1/3)

    b = 3 - 2

    b = 1

    c = 9 - 13a - 4b

    c = 9 - 13(1/3) - 4(1)

    c = 9 - 13/3 - 4

    c = 27/3 - 13/3 - 12/3

    c = 2/3

    d = 2 - a - b - c

    d = 2 - 1/3 - 1 - 2/3

    d = 6/3 - 1/3 - 3/3 - 2/3

    d = 0

    So after all that we have the following quadratic:

    f(x) = ax³ + bx² + cx + d

    f(x) = (1/3)x³ + x² + (2/3)x

    So if we substitute 2019 in for x and simplify we should get the answer we're looking for:

    f(2019) = (1/3)(2019)³ + 2019² + (2/3)(2019)

    f(2019) = (1/3)(8230172859) + 4076361 + (2/3)(2019)

    f(2019) = 2743390953 + 4076361 + 1346

    f(2019) = 2,747,468,660

    And this matches the output of a small java program that I wrote to brute-force the addition of 2019 products. This is the correct answer.

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  • 6 months ago

    T(n)=n(n+1)

    (n+2)(n+1)n-(n+1)n(n-1)

    =

    (n+1)n[n+2-n+1]

    =

    3(n+1)n

    =>

    ...2019

    SIGMA[T(n)]=

    ...n=1

    ...2019

    SIGMA[(n+2)(n+1)n-(n+1)n(n-1)]/3

    ...n=1

    =>

    (1/3)[3*2*1-0

    +4*3*2-3*2*1

    +5*4*3-4*3*2

    ----------

    2021*2020*2019-

    2020*2019*2018]

    =

    (1/3)(2021*2020*2019)

    =

    2747468660

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  • Ian H
    Lv 7
    6 months ago

    Using two standard results from sigma notation

    Σ n^2 = n(n + 1)(2n + 1)/6 = (n + 1)/6 * [2n^2 + n]

    Σn = n(n + 1)/2 = (n + 1)/6 * (3n)

    Σ n(n+ 1) = Σn^2 + Σn = (n + 1)/6 * 2(n^2 + 2n)

    Σ n(n+ 1) = n(n + 1)(n + 2)/3

    as found by Captain Matticus, LandPiratesInc and equivalent to

    llaffer’s polynomial (1/3)n^3 + n^2 + (2/3)n

    With n = 2019 the sum becomes

    (1/3) * 2019 * 2020 * 2021 = 2747468660

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  • 6 months ago

    1 × 2 + 2 × 3 + 3 × 4 + .... + 2019 × 2020

    = 2 + 6 + 12 + .... + 4078380

    sum_(n=1)^2019 n (n + 1) = 2747468660

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  • 1 * 2 + 2 * 3 + 3 * 4 + 4 * 5 + .... + 2019 * 2020

    sum(x * (x + 1) , x = 1 , x = 2019) =>

    sum(x^2 + x , x = 1 , x = 2019) =>

    sum(x^2 , x + 1 , x = 2019) + sum(x , x = 1 , x = 2019)

    The sum of the squares of the first n integer is n * (n + 1) * (2n + 1) / 6

    The sum of the first n integers is n * (n + 1) / 2

    n * (n + 1) * (2n + 1) / 6 + n * (n + 1) / 2 =>

    (1/6) * n * (n + 1) * ((2n + 1) + 3) =>

    (1/6) * n * (n + 1) * (2n + 4) =>

    (1/3) * n * (n + 1) * (n + 2)

    n = 2019

    (1/3) * 2019 * 2020 * 2021 =>

    673 * 2020 * 2021 =>

    2747468660

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  • 6 months ago

    1 x 2 = 2 = 1³/3 + 1² + 1 * 2/3

    1 x 2 + 2 x 3 = 2 + 6 = 8 = 2³/3 + 2² + 2 * 2/3

    1 x 2 + 2 x 3 + 3 x 4 = 2 + 6 + 12 = 20 = 3³/3 + 3² + 3 * 2/3

    1 x 2 + 2 x 3 + 3 x 4 + .... +2019 x 2020 = 2019³/3 + 2019² + 2019 * 2/3 = 2747468660

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