F.6 math arithmetric sequence2

用戶36732

2011-09-13 11:49 pm

Red, yellow and green squares are arranged in the following ways.

There is one red square in the 1st row from the top, three red squares in the 2nd row, five red squares in the 3rd row, and so on.

In every alternate row starting from the 2nd row, there is one yellow square between every two red squares.

In every alternate rwo starting from the 3rd row, there is one green square between every two red squares

(a) Find the no. of red squares in the 75th row.

(b) Find the no. of yellow squares in the 78th row.

(c) Find the no. of green squares in the 79th row.

回答 (1)

Solarileo

2011-09-14 12:22 am

✔ 最佳答案

(a)
no of red squares in 1st row = 1+2x0
no of red squares in 2nd row = 1+2x1
no of red squares in 3rd row = 1+2x2
.
.
.
no of red squares in nth row = 1+2(n-1)
therefore, no of red squares in 75th row = 1+2(75-1) = 149

(b)
no. of yellow squares in 2nd row = 1+2x1 - 1 = 2x1
no. of yellow squares in 4th row = 1+2x3 - 1 = 2x3
no. of yellow squares in 6th row = 1+2x5 - 1 = 2x5
...
no. of yellow squares in nth row = 2(n-1) where n is even.
therefore, no. of yellow squares in 78th row = 2(78-1) = 154

(c)
no. of green squares in 3rd row = 1+2x2 - 1 = 2x2
no. of green squares in 5th row = 1+2x4 - 1 = 2x4
no. of green squares in 7th row = 1+2x6 - 1 = 2x6
...
no. of yellow squares in nth row = 2(n-1) where n is odd.
therefore, no. of green squares in 79th row = 2(79-1) = 156

參考: myself

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