Question

Can someone please help me​

Answer 1

Answer:

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 squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

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