hard sum

This series creates a series of triangular numbers:

 

1  =  1

1 + 2  =  3

1 + 2 + 3  =  6

1 + 2 + 3 + 4  =  10

...

1 + 2 + 3 + ... + 2020  =     < which I don't need to calculate, because I won't need it >

 

The formula for the sum of n triangular numbers is:  Sum  =  n · (n + 1) · (n + 2) / 6

                                   

Sum  =  2020 · (2020 + 1) · (2020 + 2) / 6  =  1 375 775 540

geno3141 misread the question !! I think what the poster wants is this:

 

n · (n + 1) · (n + 2) / 6 =2020, solve for n

 

n = ~ 22. But then with 22, the total =2024 NOT 2020.