BG FORMULA
A grid has a
large number of squares hidden in them. Have a look at this:
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For
m<n;
TNoS =
Sm2
+ (n-m) Sm
NOTE:
It is important that in any grid the lower number ( whether it is column or row
) is substituted in the position of m and the higher number is substituted in
the position of n.
EXAMPLES
#1
The
following grid has 2(m) columns and 2(n) rows.
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You can
count for yourself!! (The four smaller squares and the one big outer square)
#2
The
following grid has 10(n) columns and 1(m) row.
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Hence the
total number of squares present in the grid is = S12
+ (10-1) S1
= ( 1 ) + 9 = 10.
You can count for yourself again!! (There are only 10 squares here and the other figures you can think of would be rectangles!!)
#3
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Hence the
total number of squares present in the grid is =
S72
+ (10-7) S7=
( 1+4+9+16+25+36+49) + 3 * (1+2+3+4+5+6+7) = 224.
Try counting
that!!. (First count all the smaller squares made of 1 block - 70. Then the
squares made of 4 blocks, then squares made of 9 blocks and so on…)
PS:
I DO NOT claim to be the FIRST or ONLY person to have discovered this formula.
There maybe others who have proposed this formula before, but I am unaware of
any such person as yet.
For further clarification e-mail me at blessongregory@gmail.com