Bài toán 20: Tính tổng $S = sec\frac{\pi }{{13}} + sec\frac{{3\pi }}{{13}} + sec\frac{{5\pi }}{{13}} + sec\frac{{7\pi }}{{13}} + sec\frac{{9\pi }}{{13}} + sec\frac{{11\pi }}{{13}}$.
Trong đó ký hiệu $\sec x=\frac{1}{\cos x}$.
Hãy tổng quát hóa bài toán.
Lời giải bài toán 20:
Đặt ${z_k} = \cos \frac{{(2k - 1)\pi }}{{13}} + {\rm{i}}{\rm{.}}\sin \frac{{({\rm{2k - 1}})\pi }}{{{\rm{13}}}}$ với ${\rm{k = 1}},{\rm{2}},...,{\rm{13}}$ là các nghiệm phức của PT :
${{\rm{z}}^{{\rm{13}}}}{\rm{ + 1 = }}({\rm{z + 1}}).({{\rm{z}}^{{\rm{12}}}}{\rm{ - }}{{\rm{z}}^{{\rm{11}}}}{\rm{ + }}{{\rm{z}}^{{\rm{10}}}}{\rm{ - }}{{\rm{z}}^{\rm{9}}}{\rm{ + }}{{\rm{z}}^{\rm{8}}}{\rm{ - }}{{\rm{z}}^{\rm{7}}}{\rm{ + }}{{\rm{z}}^{\rm{6}}}{\rm{ - }}{{\rm{z}}^{\rm{5}}}{\rm{ + }}{{\rm{z}}^{\rm{4}}}{\rm{ - }}{{\rm{z}}^{\rm{3}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ - z + 1}}){\rm{ = 0}}$
$ \Leftrightarrow (z + 1).{z^6}.\left( {{z^6} - {z^5} + {z^4} - {z^3} + {z^2} - z + 1 - \frac{1}{z} + \frac{1}{{{z^2}}} - \frac{1}{{{z^3}}} + \frac{1}{{{z^4}}} - \frac{1}{{{z^5}}} + \frac{1}{{{z^6}}}} \right) = 0$
Đặt ${z_k} + \frac{1}{{{z_k}}} = 2\cos \frac{{(2k - 1)\pi }}{{13}} = {x_k}$,với $k = 1,2,...,6$ thì:
$z + \frac{1}{z} = x$
${z^2} + \frac{1}{{{z^2}}} = {\left( {z + \frac{1}{z}} \right)^2} - 2 = {x^2} - 2$
${z^3} + \frac{1}{{{z^3}}} = {\left( {z + \frac{1}{z}} \right)^3} - 3\left( {z + \frac{1}{z}} \right) = {x^3} - 3x$
${z^4} + \frac{1}{{{z^4}}} = {\left( {z + \frac{1}{z}} \right)^4} - 4\left( {{z^2} + \frac{1}{{{z^2}}}} \right) - 6 = {x^4} - 4({x^2} - 2) - 6 = {x^4} - 4{x^2} + 2$
${z^5} + \frac{1}{{{z^5}}} = {\left( {z + \frac{1}{z}} \right)^5} - 5\left( {{z^3} + \frac{1}{{{z^3}}}} \right) - 10\left( {z + \frac{1}{z}} \right) = {x^5} - 5({x^3} - 3x) - 10x = {x^5} - 5{x^3} + 5x$
$\begin{array}{l}{z^6} + \frac{1}{{{z^6}}} = {\left( {z + \frac{1}{z}} \right)^6} - 6\left( {{z^4} + \frac{1}{{{z^4}}}} \right) - 15\left( {{z^2} + \frac{1}{{{z^2}}}} \right) - 20\\= {x^6} - 6[{x^4} - 4({x^2} - 2) - 6] - 15({x^2} - 2) - 20 = {x^6} - 6{x^4} + 9{x^2} - 2\end{array}$
Vậy thì:
${z^6} - {z^5} + {z^4} - {z^3} + {z^2} - z + 1 - \frac{1}{z} + \frac{1}{{{z^2}}} - \frac{1}{{{z^3}}} + \frac{1}{{{z^4}}} - \frac{1}{{{z^5}}} + \frac{1}{{{z^6}}}$
$ = ({x^6} - 6{x^4} + 9{x^2} - 2) - ({x^5} - 5{x^3} + 5x) + ({x^4} - 4{x^2} + 2) - ({x^3} - 3x) + ({x^2} - 2) - x + 1$
$ = {x^6} - {x^5} - 5{x^4} + 4{x^3} + 6{x^2} - 3x - 1 = \sum\limits_{k = 0}^6 {{a_k}} {x^k} = 0$
Từ đó ta tính được :
$$\begin{array}{rcl}S &=& \sum\limits_{k = 1}^6 {\sec } \frac{{(2k - 1)\pi }}{{13}}\\&=& \frac{{\sum\limits_{j = 1}^6 {{2^{ - 5}}} \prod\limits_{k = 1,k \ne j}^6 {{x_k}} }}{{{2^{ - 6}}\prod\limits_{k = 1}^6 {{x_k}} }}\\&=& \frac{{2\cdot( - {a_1})}}{{{a_0}}} = \frac{{2\cdot3}}{{ - 1}} = \boxed{\displaystyle - 6}\end{array}$$
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Đề mới:
Bài toán 22: Cho ${a_k} = 1 + \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{{2k - 1}}$ với ${\rm{k}} = {\rm{1}},{\rm{2}},...,{\rm{n}}$
Tính tổng $\frac{1}{2}a_n^2 + {({a_n} - {a_1})^2} + {({a_n} - {a_2})^2} + ... + {({a_n} - {a_{n - 1}})^2}$
Bài toán 23: Tính tổng $\sum\limits_{k = 0}^{n - 1} {{{\cos }^{ - 1}}} \left( {\frac{{{n^2} + {k^2} + k}}{{\sqrt {{n^4} + {k^4} + 2{k^3} + 2{n^2}{k^2} + 2{n^2}k + {n^2} + {k^2}} }}} \right)$