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

 

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civonamzuk  May 31, 2015

Best Answer 

 #1
avatar+18715 
+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: 

2+0 Answers

 #1
avatar+18715 
+10
Best Answer

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
avatar+78754 
+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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