+1693 How many turquoise squares will be required to build the twentieth figure in this pattern?
+26397 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 = (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 $}}
+26397 How many turquoise squares will be required to build the twentieth figure in this pattern?
n = turquoise squares
n-1 = white squares
n + n - 1 = all squares = (2i−1)2 i is the number of the figure
\begin{array}{rcl} n+n-1 &=& (2i-1)^2\\ 2n-1 &=& (2i-1)^2\\ 2n &=& 1+(2i-1)^2\\\\ n &=& \dfrac{1+(2i-1)^2}{2}\\ \end{array}\\\\\\ \boxed{~n_i = \dfrac{1+(2i-1)^2}{2} \qquad \rm{or} \qquad$ 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 $}}
+130477 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.....
