Sum of the 66 consecutive integers is 5181, then sum of the squares is?

• sum of 66 consecutive integers is 5181,
  Let consecutive integers are : x , (x + 1) , ( x + 2) , ( x + 3) ........ (x + 65)
  
  now, sum of consecutive integers = x + (x + 1) + (x + 2) + (x + 3) + (x + 4) ....... + (x + 65)
  = 66x + (1 + 2 + 3 + .... + 65)
  = 66x + 65(65 + 1)/2
  = 66x + 65 × 33
  
  so, 66x + 65 × 33 = 5181
  11(6x + 65 × 3) = 471 × 11
  (6x + 65 × 3) = 471
  6x = 471 - 195 = 276
  x = 46
  
  hence, consecutive integers are 46, 47, 48, ... 111
  
  now, sum of square of consecutive integers
  = 46² + 47² + 48² + .... 111²
  =
  use formula
  
  =
  
  = 430,661
• 4 years agoHelpfull: Yes(0) No(0)