Given, $${a_k} = \cos \left( {{{k\pi } \over 7}} \right) + i\sin \left( {{{k\pi } \over 7}} \right) = {e^{{{k\pi } \over 7}i}}$$

We have to find $${{\sum\limits_{k = 1}^{12} {\left| {{a_{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right|} }}$$

$${a_{k + 1}} = \cos \left( {{{k + 1} \over 7}} \right)\pi + i\sin \left( {{{k + 1} \over 7}} \right)\pi = {e^{i\left( {{{k + 1} \over 7}} \right)\pi }}$$

$$\therefore$$ $${a_{k + 1}} - {a_k} = {e^{\left( {{{k + 1} \over 7}} \right)\pi i}} - {e^{{{k\pi } \over 7}i}}$$

$$ = {e^{{{k\pi } \over 7}i}}\,.\,\,{e^{{\pi \over 7}i}} - {e^{{{k\pi } \over 7}i}}$$

$$ = {e^{{{k\pi } \over 7}i}}\left( {{e^{{\pi \over 7}i}} - 1} \right)$$

$$\therefore$$ $$\left| {{a_{k + 1}} - {a_k}} \right| = \left| {{e^{{{k\pi } \over 7}i}}\left( {{e^{{\pi \over 7}i}} - 1} \right)} \right|$$

$$ = \left| {{e^{{{k\pi } \over 7}i}}} \right|\left| {{e^{{\pi \over 7}i}} - 1} \right|$$

$$ = \left| {{e^{{\pi \over 7}i}} - 1} \right|$$

If $$z = {e^{i\theta }} = \cos \theta + i\sin \theta $$

then $$\left| z \right| = \sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } = 1$$

that is why $$\left| {{e^{{{k\pi } \over 7}i}}} \right| = 1$$

Now, $${a_{4k - 1}} = {e^{\left( {{{4k - 1} \over 7}} \right)\pi i}}$$

$${a_{4k - 2}} = {e^{\left( {{{4k - 2} \over 7}} \right)\pi i}}$$

$${a_{4k - 1}} - {a_{4k - 2}} = {e^{\left( {{{4k - 1} \over 7}} \right)\pi i}} - {e^{\left( {{{4k - 2} \over 7}} \right)\pi i}}$$

$$ = {e^{{{4k\pi } \over 7}i}}\,.\,{e^{ - {\pi \over 7}i}} - {e^{{{4k\pi } \over 7}}}\,.\,{e^{ - {{2\pi } \over 7}i}}$$

$$ = {e^{{{4k\pi } \over 7}i}}\left( {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right)$$

$$\therefore$$ $$\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right| = \left| {{e^{{{4k\pi } \over 7}i}}} \right|\left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|$$

$$ = \left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|$$

Now, $${{\sum\limits_{k = 1}^{12} {\left| {{a_{k + 1}} - {a_k}} \right|} } \over {\sum\limits_{k = 1}^3 {\left| {{a_{4k - 1}} - {a_{4k - 2}}} \right|} }}$$

$$ = {{12\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {3\left| {{e^{ - {\pi \over 7}i}} - {e^{ - {{2\pi } \over 7}i}}} \right|}}$$

$$ = 4\,.\,{{\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {\left| {\left( {{e^{ - {\pi \over 7}i}}1 - {e^{ - {\pi \over 7}i}}} \right)} \right|}}$$

$$ = 4\,.\,{{\left| {{e^{{\pi \over 7}i}} - 1} \right|} \over {\left| {1 - {e^{ - {\pi \over 7}i}}} \right|}}$$

$$ = 4\,.\,{{\left| {{e^{{\pi \over 7}i}}\left( {1 - {e^{ - {\pi \over 7}i}}} \right)} \right|} \over {\left| {1 - {e^{ - {\pi \over 7}i}}} \right|}}$$

$$ = 4\,.\,\left| {{e^{{\pi \over 7}i}}} \right|$$

= 4 as $$\left| {{e^{{\pi \over 7}i}}} \right| = 1$$