Sum the series
1*2+2*3+3*4+ ... +2019*2020
General term = n(n+1) for n ∈ N
We want:
∑ n(n+1) for n=1 to 2019
= ∑ n^2 + n
= ∑ n^2 + ∑ n
= n(n+1)(2n+1)/6 + n(n+1)/2 , replace n with 2019 and evaluate
To find the sum of the series 1*2 + 2*3 + 3*4 + ... + 2019*2020, we can use the formula for the sum of an arithmetic series.
The series can be written as:
(1*2) + (2*3) + (3*4) + ... + (2019*2020)
To determine the common difference, we subtract each term from the one that follows it:
2 - 1 = 1
3 - 2 = 1
4 - 3 = 1
...
2020 - 2019 = 1
We can see that the common difference between each term is 1.
Let's denote the first term of the series as a1 = 1*2 and the last term as an = 2019*2020.
Using the formula for the sum of an arithmetic series, the sum S can be calculated as:
S = (n/2)(a1 + an)
where n is the number of terms in the series.
In this case, the number of terms n can be determined by finding the common difference and the first and last terms. We have:
a1 = 1*2 = 2
d = 1
an = 2019*2020
We can calculate the number of terms n using the formula:
(an - a1)/d + 1
n = (2019*2020 - 2)/1 + 1
n = 4078499
Now we have all the required values to calculate the sum of the series:
S = (n/2)(a1 + an)
S = (4078499/2)(2 + an)
Substituting an = 2019*2020, we have:
S = (4078499/2)(2 + 2019*2020)
Calculating this expression will give us the sum of the series.
Note: Due to the large value of n and the calculation involved, it might not be practical to compute the exact sum manually. An alternative approach could be to use a computer program or a calculator that supports large calculations.