square 20 has 44 green squares
square 21 has 45 green squares
Step-by-step explanation:
To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).
For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.
The side length is 3 for square 1, 4 for square 2, and so on.
Let
n= square number (1, 2,3...)
L = side length (3,4,5...)
G1(n) = function that gives the number of green squares for square n, n=odd
G2(n) = function that gives the number of green squares for square n, n=even
side length, L=n+2 (1)
G1(n) = twice the side length less one, as discussed above
G1(n) = 2L-1 now substitute L=n+2
G1(n) = 2(n+2) -1 simplify
G1(n) = 2n + 3
Check:
for n=1, square 1 has 2*1+3 = 5 green squares ... checks
for n=3, square 3 has 2*3+3 = 9... checks
for n=5, square 5 has 2*5+3 = 13 checks
For even squares, it is even easier, because
G2(n) = 2L = 2(n+2)
check:
for n=2, square 2 has 2(2+2) = 8 green squareschecks
for n=4, square 4 has 2(4+2) = 12 green squareschecks.
Fincally, we apply our formula to n=20 and n=21
square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars
square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares
