Assume A ≤ B
2006-07-18 12:05:36 · 8 answers · asked by Scott R 6 in Science & Mathematics ➔ Mathematics
the A+1 lines are all parallel.
the B+1 lines are all parallel.
the spacings between all adjacent lines are the same
2006-07-18 12:07:48 · update #1
The answer is NOT A*B, thats just the number of squares of side length 1.
2006-07-18 12:11:25 · update #2
tricialou68:
was there ever a mathematics question you did care about?
go somewhere else if you dont like it, no one forced you to answer.
2006-07-18 12:15:55 · update #3
Note to Novice:
Your analysis is correct so far, but what IS that sum (in simplest form in terms of A and B of course).
2006-07-18 12:46:33 · update #4
Ray D, get a life yourself. Preferably one in which you have some mathematical ability if you're going to answer such questions.
2006-07-18 17:15:41 · update #5
Based on Novice's good work, here are the sums.
sum of (A-i)*(B-i), i goes from 0 to A-1
=
sum of (AB - iA - iB + i^2), i goes from 0 to A-1
=
AB(A) - (A+B)(sum of i from 0 to A-1) + (sum of i^2, i from 0 to A-1)
=
A^2*B - (A+B)( (A-1)*(A)/2 ) + ( (A-1)(A)(2A-1)/6 )
I'll let you simplify this if you like.
Notice that A*B repeatedly added for i = 0 to i = A-1 is A*B*A.
The formula for the sum of the first n counting numbers is
sum of i = n(n+1)/2.
Also, the formula for the sum of the squares of the first n counting numbers is
sum of i*i = (n)(n+1)(2n+1)/6
Above I substituted A-1 for n in both summation formulae.
2006-07-18 17:14:18 · answer #1 · answered by Johnny 2 · 1⤊ 2⤋
A times B.
2006-07-18 12:08:19 · answer #2 · answered by zee_prime 6 · 0⤊ 0⤋
The answer is
A*B + (A-1)*(B-1) + (A-2)*(B-2) + ...
or (since I don't know how to use such symbols as "SUM" using my keyboard :-))
Sum of (A-i)*(B-i), with i goes from 0 to A-1.
Reason:
A*B = number of squares with size 1
(A-1)*(B-1) = number of squares with size 2
(A-2)*(B-2) = number of squares with size 3
...
(A-(A-1))*(B-(A-1)) = number of squares with size A
(all these assuming that A <= B)
2006-07-18 12:34:12 · answer #3 · answered by Novice 4 · 0⤊ 0⤋
I see some good work has already been done, so I'll just suggest the more general equivalent question for n-space.
How many n-dimensional hypercubes in an A1 x A2 x A3 x....x An configuration?
2006-07-18 18:15:20 · answer #4 · answered by Jimbo 5 · 0⤊ 0⤋
3 attempts, 3 failures. Still thinking.
2006-07-18 12:09:37 · answer #5 · answered by lazwatson 3 · 0⤊ 0⤋
AB... now go get a life
2006-07-18 12:08:45 · answer #6 · answered by Anonymous · 0⤊ 0⤋
A*B sounds right.
2006-07-18 12:09:53 · answer #7 · answered by Tony Walls 3 · 0⤊ 0⤋
WOW!!! Does anyone REALLY CARE???
2006-07-18 12:12:38 · answer #8 · answered by tricialou68 1 · 0⤊ 0⤋