+0  
 
0
91
2
avatar

Compute \(1^2 + 3^2 + 5^2 + \dots + 99^2\)

 Jul 12, 2020
 #1
avatar
0

Use "Sum of consecutive squares" formula to sum them up:

 

n =99; a = 1/6 * (2*n + 1)*n*(n + 1) =328,350

 Jul 12, 2020
 #2
avatar+31085 
+1

The numbers aren't consecutive, they increase by 2, so:

 

 Jul 13, 2020

57 Online Users

avatar
avatar
avatar
avatar
avatar