Let n be a positive integer. Let

$$A = \sum\limits_{k = 0}^n {{{( - 1)}^k}{}^n{C_k}\left[ {{{\left( {{1 \over 2}} \right)}^k} + {{\left( {{3 \over 4}} \right)}^k} + {{\left( {{7 \over 8}} \right)}^k} + {{\left( {{{15} \over {16}}} \right)}^k} + {{\left( {{{31} \over {32}}} \right)}^k}} \right]} $$. If

$63A = 1 - {1 \over {{2^{30}}}}$, then n is equal to _____________.

$$A = \sum {{{( - 1)}^k}{}^n{C_k}{{\left( {{1 \over 2}} \right)}^k}} + \sum {{{( - 1)}^k}{}^n{C_k}{{\left( {{3 \over 4}} \right)}^k}} + .....$$<br><br>$$ = {\left( {1 - {1 \over 2}} \right)^n} + {\left( {1 - {3 \over 4}} \right)^n} + ..... + {\left( {1 - {{31} \over {32}}} \right)^n}$$<br><br>$$ = {\left( {{1 \over 2}} \right)^n} + {\left( {{1 \over 2}} \right)^{2n}} + {\left( {{1 \over 2}} \right)^{3n}} + ..... + {\left( {{1 \over 2}} \right)^{5n}}$$<br><br>$$ = {\left( {{1 \over 2}} \right)^n}\left( {{{1 - {{\left( {{1 \over 2}} \right)}^{5n}}} \over {1 - {{\left( {{1 \over 2}} \right)}^n}}}} \right) = {{{2^{5n}} - 1} \over {{2^{5n}}({2^n} - 1)}}$$ <br><br>$\therefore$ $$63A = {{63\left( {{2^{5n}} - 1} \right)} \over {{2^{5n}}\left( {{2^n} - 1} \right)}}$$ = $${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$$ <br><br>Given, $63A = 1 - {1 \over {{2^{30}}}}$ <br><br>$\therefore$ $${{63} \over {\left( {{2^n} - 1} \right)}}\left( {1 - {1 \over {{2^{5n}}}}} \right)$$ = $1 - {1 \over {{2^{30}}}}$ <br><br>For n = 6, L.H.S = R.H.S <br><br>$\therefore$ n = 6