The following table gives the values of ${x_n}$ and ${A_n}$, for $1 \leqslant n \leqslant 3$. Copy and complete the table. (Do not write on this page.)

The process described above is repeated. Find ${A_6}$.

Consider an initial square of side length $k {\text{ cm}}$. The process described above is repeated indefinitely. The total area of the shaded regions is $k {\text{ c}}{{\text{m}}^2}$. Find the value of $k$.

valid method for finding side length     (M1)

eg   ${8^2} + {8^2} = {c^2},{\text{ }}45 - 45 - 90{\text{ side ratios, }}8\sqrt 2 ,{\text{ }}\frac{1}{2}{s^2} = 16,{\text{ }}{x^2} + {x^2} = {8^2}$

correct working for area     (A1)

eg   $\frac{1}{2} \times 4 \times 4$

     A1A1     N2N2

[4 marks]

METHOD 1

recognize geometric progression for ${A_n}$     (R1)

eg   ${u_n} = {u_1}{r^{n - 1}}$

$r = \frac{1}{2}$     (A1)

correct working     (A1)

eg   $32{\left( {\frac{1}{2}} \right)^5};{\text{ 4, 2, 1, }}\frac{1}{2},{\text{ }}\frac{1}{4},{\text{ }} \ldots$

${A_6} = 1$     A1     N3

METHOD 2

attempt to find ${x_6}$     (M1)

eg   $8{\left( {\frac{1}{{\sqrt 2 }}} \right)^5},{\text{ }}2\sqrt 2 ,{\text{ 2, }}\sqrt 2 ,{\text{ 1, }} \ldots$

${x_6} = \sqrt 2$     (A1)

correct working     (A1)

eg   $\frac{1}{2}{\left( {\sqrt 2 } \right)^2}$

${A_6} = 1$     A1     N3

[4 marks]

METHOD 1

recognize infinite geometric series     (R1)

eg   ${S_n} = \frac{a}{{1 - r}},{\text{ }}\left| r \right| < 1$

area of first triangle in terms of $k$     (A1)

eg   $\frac{1}{2}{\left( {\frac{k}{2}} \right)^2}$

attempt to substitute into sum of infinite geometric series (must have $k$)     (M1)

eg   $\frac{{\frac{1}{2}{{\left( {\frac{k}{2}} \right)}^2}}}{{1 - \frac{1}{2}}},{\text{ }}\frac{k}{{1 - \frac{1}{2}}}$

correct equation     A1

eg   $\frac{{\frac{1}{2}{{\left( {\frac{k}{2}} \right)}^2}}}{{1 - \frac{1}{2}}} = k,{\text{ }}k = \frac{{\frac{{{k^2}}}{8}}}{{\frac{1}{2}}}$

correct working     (A1)

eg   ${k^2} = 4k$

valid attempt to solve their quadratic     (M1)

eg   $k(k - 4),{\text{ }}k = 4{\text{ or }}k = 0$

$k = 4$     A1     N2

METHOD 2

recognizing that there are four sets of infinitely shaded regions with equal area     R1

area of original square is ${k^2}$     (A1)

so total shaded area is $\frac{{{k^2}}}{4}$     (A1)

correct equation $\frac{{{k^2}}}{4} = k$     A1

${k^2} = 4k$    (A1)

valid attempt to solve their quadratic     (M1)

eg   $k(k - 4),{\text{ }}k = 4{\text{ or }}k = 0$

$k = 4$     A1     N2

[7 marks]