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# How many turquoise squares will be required to build the twentieth figure in this pattern?

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How many turquoise squares will be required to build the twentieth figure in this pattern?

civonamzuk  May 31, 2015

#1
+18829
+10

How many turquoise squares will be required to build the twentieth figure in this pattern?

n = turquoise squares

n-1 = white squares

n + n - 1  = all squares = $$\small{\text{  (2i-1)^2 }}$$
i is the number of the figure

$$\begin{array}{rcl} n+n-1 &=& (2i-1)^2\\ 2n-1 &=& (2i-1)^2\\ 2n &=& 1+(2i-1)^2\\\\ n &=& \dfrac{1+(2i-1)^2}{2}\\ \end{array}\\\\\\ \boxed{~n_i = \dfrac{1+(2i-1)^2}{2} \qquad \rm{or} \qquad n_i = 1 + 2i(i-1)~}\\\\ n_1 = \dfrac{1+(2*1-1)^2}{2} = 1\\\\ n_2 = \dfrac{1+(2*2-1)^2}{2} = 5\\\\ n_3 = \dfrac{1+(2*3-1)^2}{2} = 13\\\\ n_4 = \dfrac{1+(2*4-1)^2}{2} = 25\\\\ \cdots\\\\ n_{20} = \dfrac{1+(2*20-1)^2}{2} = 761 \small{\text{\qquad \rm{or} \qquad n_{20}= 1+2\cdot 20(20-1) = 1+40\cdot 19=761 }}$$

heureka  May 31, 2015
Sort:

#1
+18829
+10

How many turquoise squares will be required to build the twentieth figure in this pattern?

n = turquoise squares

n-1 = white squares

n + n - 1  = all squares = $$\small{\text{  (2i-1)^2 }}$$
i is the number of the figure

$$\begin{array}{rcl} n+n-1 &=& (2i-1)^2\\ 2n-1 &=& (2i-1)^2\\ 2n &=& 1+(2i-1)^2\\\\ n &=& \dfrac{1+(2i-1)^2}{2}\\ \end{array}\\\\\\ \boxed{~n_i = \dfrac{1+(2i-1)^2}{2} \qquad \rm{or} \qquad n_i = 1 + 2i(i-1)~}\\\\ n_1 = \dfrac{1+(2*1-1)^2}{2} = 1\\\\ n_2 = \dfrac{1+(2*2-1)^2}{2} = 5\\\\ n_3 = \dfrac{1+(2*3-1)^2}{2} = 13\\\\ n_4 = \dfrac{1+(2*4-1)^2}{2} = 25\\\\ \cdots\\\\ n_{20} = \dfrac{1+(2*20-1)^2}{2} = 761 \small{\text{\qquad \rm{or} \qquad n_{20}= 1+2\cdot 20(20-1) = 1+40\cdot 19=761 }}$$

heureka  May 31, 2015
#2
+81032
+5

I used a slightly different approach from heureka's.......we still get the same answers....!!!

1, 5, 13, 25.....

It appears that the series is defined by ......

1 = 1 +  0   =  1 +  4(0)

5 = 1 +  4   =  1 + 4(1)

13 = 1 + 12  = 1 +  4(3)

25 = 1 + 24  =  1 + 4(6)

41= 1 + 40   = 1 + 4(10)

So, the nth total is defined as........

1 + 4[sum of the first n whole numbers] =

1 + 4[(n)(n - 1)/2]  =

1 + 2[(n)(n-1] =

So....the 20th figure will have  ...

1 + 2[(20)(19)]   =  1 + 2[380] = 1 + 760 = 761 turquoise squares.....

CPhill  May 31, 2015

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