find the remainder: 1^2+2^2+3^2+................+100^2/4

• we know the formula for summation of square of n numbers = [ n * (n+1) * (2n+1)] / 6
  here, n= 100
  so, sum= [ 100 * 101 * 201 ] / 6
  = 338350
  so, last two digit is 50 hence by dividing (50 / 4)
  we get remainder 2.
  answer is 2.
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• use formula for sum of square of first 'n' natural numbers.
  Answer is:
  84587.5
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• The Answer is 84587.5 if the Question is (1^2+2^2+3^2+....+100^2)/4
  
  The Answer is 330850 if the question is 1^2+2^2+3^2+....+(100^2)/4
  
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• Remainder is 2.
  
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