find the sum of all the perfect sqaes lying between 4000 and 7000

• perfect squares between 4000 and 7000 are 64^2 = 4096 , 65^2, ..... 83^2
  sum of squares = n(n+1)(2n+1)/6
  =>(1^2 + 2^2 + 3^2 + 4^2 + ....... + 83^2) - (1^2 + 2^2 + 3^2 + 4^2 + ....... +63^2)
  
  => {(83)(84)(167)-(63)(64)(127)}/6
  =>108710
• 10 years agoHelpfull: Yes(4) No(0)
• 108710
  
  perfect squares between 4000 and 7000
  64^2,65^2,66^2.....83^2
  
  sum = 64^2 + 65^2 + ......82^2 + 83^2
  = (1^2+2^2+3^2+....+83^2)-(1^2+2^2+....+63^2)
  = (83*84*167/6)-(63*64*127/6) [sum of squares = n*(n+1)*(2n+1)/6 ]
  =194054 - 85344
  =108710
  
• 10 years agoHelpfull: Yes(2) No(0)
• perfect sqr just greater than 4000 is 64^2=4096
  perfect sqr just less than 7000 is 83^2=6889
  so find the sum of the numbers starting from 64 to 83.
  64*20 + sum(1 to 19)
  =1280+190
  =1470
• 10 years agoHelpfull: Yes(0) No(3)