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

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.

• 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

• 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

• 6 months ago

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

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

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

• 6 months ago

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