+0  
 
0
70
1
avatar

The sum 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6.

 

What is the value of 21^2 + 22^2 + 23^2 + ... + 40^2 + 41^2 + 42^2?

 Jun 3, 2021
 #1
avatar+121054 
+1

The  sum  =     sum of all the squares  1 through  42    minus  the  sum  of all the squares  1   through  20

 

So  we  have

 

42 (43) (85)/6    -  20 (21)(41)/6      =

 

22715

 

 

cool cool cool

 Jun 3, 2021

35 Online Users