+0  
 
0
81
4
avatar

Compute 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ...  + n) for n = 2020.

 Jun 1, 2020
 #1
avatar+21953 
0

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

 Jun 1, 2020
 #2
avatar
0

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.

 Jun 1, 2020
 #3
avatar+40 
+1

No geno3141 is correct, the problem states to compute the expression when n = 2020, not when the expression equals 2020.

AWQSed  Jun 1, 2020
 #4
avatar
0

Oh, OK. Thanks for that. I misread the question.

Guest Jun 1, 2020

32 Online Users

avatar
avatar
avatar
avatar