how many total no. of square in chess board...?

• no. of square of size 1*1=8*8=8^2
  no. of square of size 2*2=7*7=7^2
  no. of square of size 3*3=6*6=6^2
  ........
  similarly
  no of square of size 8*8=1*1
  so total no. of square=
  1^2+2^2+3^2+...+8^2=8(8+1)(2*8+1)/6=204
• 11 years agoHelpfull: Yes(37) No(0)
• n(n+1)(2n+1)/6
  where n= no rows or clm
  so 8(8+1)(16+1)/6=204
• 11 years agoHelpfull: Yes(13) No(1)
• 1^2+2^2+....+8^2 = 204
• 11 years agoHelpfull: Yes(4) No(0)
• answer should be 204 -:)
• 11 years agoHelpfull: Yes(4) No(1)
• Consider the lefthand vertical edge of a square of size 1 x 1.
  This edge can be in any one of 8 positions. Similarly, the top
  edge can occupy any one of 8 positions for a 1 x 1 square.
  So the total number of 1 x 1 squares = 8 x 8 = 64.
  
  For a 2 x 2 square the lefthand edge can occupy 7 positions and
  the top edge 7 positions, giving 7 x 7 = 49 squares of size 2 x 2.
  Continuing in this way we get squares of size 3 x 3, 4 x 4 and so on.
  the answer turns to be 204
• 11 years agoHelpfull: Yes(3) No(0)
• for num of squares formula is summation of n^2 where n is num of rows or columns
  for num of rectangles formula is summation of n^3 where n is num of rows or colomns
• 11 years agoHelpfull: Yes(3) No(0)
• 1^2+2^2+3^2+4^2+5^2+6^2+7^2+8^2
  use this formula
  n(n+1)(2n+1)/6 where n=8
  =8(8+1)(2*8+1)/6=204
• 11 years agoHelpfull: Yes(3) No(0)
• no of squares = 1^2 + 2^2 +...+8^2 = 204
• 11 years agoHelpfull: Yes(3) No(1)
• toal no. of square=1^2+2^+.....+8^2=204
• 11 years agoHelpfull: Yes(1) No(0)
• ANS 204
  simple formula:
  number of squares in s square of side n*n=1^2+2^2+...+n^2
  now, chess board is a square of side 8*8
  so, no of squares=1^2+2^2+....+8^2=204
• 11 years agoHelpfull: Yes(1) No(0)
• In ques it is not said that repetition is allowed or not.
  1: If repetition is alowed--given solution is true.
  2: if not...then the answer will be: 8^2+4^2+2^2+2^2+1+1+1+1=92
• 11 years agoHelpfull: Yes(0) No(3)
• answer is 204
  
• 11 years agoHelpfull: Yes(0) No(0)
• 204 !!!!
• 11 years agoHelpfull: Yes(0) No(0)
• here is an way to find it-
  let n=no of side horizontally
  x=no of side vertically
  so the no of squares will be=(n*x)+(n-1*x-1)+(n-2*x-2)+ .....
  If n>x then stop when n=1
  If x>n then stop when x=1
  but here n=x so,
  (8*8)+(7*7)+(6*6)+(5*5)+(4*4)+(3*3)+(2*2)+(1*1)=204
• 11 years agoHelpfull: Yes(0) No(0)