گام به گام تمرین صفحه 6 درس 1 حسابان (1) (جبر و معادله)

\(1 + 3 + 5 + ... + (2n - 1) = {n^2}\)

\(\left. \begin{array}{l}{a_1} = 1\\\\{a_n} = 2n - 1\end{array} \right\} \Rightarrow {S_n} = \frac{n}{2}({a_1} + {a_n}) = \frac{n}{2}(1 + 2n - 1) = {n^2}\)

\(\begin{array}{l}\left. \begin{array}{l}{a_1} = 102\\\\{a_n} = 996\\\\d = \frac{{{a_n} - {a_{}}}}{{n - 1}}\end{array} \right\} \Rightarrow 6 = \frac{{996 - 102}}{{n - 1}} \Rightarrow n = \frac{{894}}{6} + 1 = 150\\\\ \Rightarrow {S_{150}} = \frac{n}{2}({a_1} + {a_n}) = \frac{{150}}{2}(102 + 996) = 82,350\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}a = 5\\\\d = 3\end{array} \right\} \Rightarrow {a_n} = a + (n - 1)d = 3n + 2\\\\{S_n} > 493 \Rightarrow n = ?\\\\{S_n} = \frac{n}{2}({a_1} + {a_n}) = \frac{n}{2}(3n + 7) > 493\\\\ \Rightarrow 3{n^2} + 7n - 986 > 0 \Rightarrow n > 17 \Rightarrow n = 18\\\\ \Rightarrow {S_{18}} = 594 > 493\end{array}\)

\(\begin{array}{l}{S_n} = \frac{n}{2}[2a + (n - 1)d]\\\\150 + 135 = \frac{{20}}{2}[2a + 19d] \Rightarrow 20a + 190d = 285\\\\ \Rightarrow 4a + 38d = 57\,\,\,(1)\\\\a + (a + 2d) + (a + 4d) + ... + (a + 18d) = 135\\\\ \Rightarrow 10a + 90d = 135 \Rightarrow 2a + 18d = 27\,\,\,(2)\\\\(1)\,,\,(2):\\\\\left\{ \begin{array}{l}4a + 38d = 57\\\\2a + 18d = 27\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 0\\\\d = 1/5\end{array} \right.\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}{a_1} = {2^{ - 1}} = {2^0} = 1\\\\q = 2\\\\{S_n} = 255 \Rightarrow n = ?\end{array} \right\} \Rightarrow {S_n} = {a_1}\frac{{1 - {q^n}}}{{1 - q}}\\\\ \Rightarrow 255 = 1 \times \frac{{1 - {2^n}}}{{1 - 2}} \Rightarrow {2^n} - 1 = 255 \Rightarrow {2^n} = 256 = {2^8}\\\\ \Rightarrow n = 8\end{array}\)

\(\begin{array}{l}\frac{1}{2}\,,\,\frac{1}{4}\,,\,\frac{1}{8}\,,\,...\\\\\left. \begin{array}{l}a = \frac{1}{2}\\\\q = \frac{1}{2}\\\\{S_n} > \frac{{99}}{{100}} \Rightarrow {n_{\min }} = ?\end{array} \right\} \Rightarrow {S_n} = a\frac{{1 - {q^n}}}{{1 - q}} = \frac{1}{2}\frac{{1 - {{(\frac{1}{2})}^n}}}{{1 - \frac{1}{2}}}\\\\ \Rightarrow {S_n} = 1 - {(\frac{1}{2})^n} > \frac{{99}}{{100}} \Rightarrow {(\frac{1}{2})^n} < 1 - \frac{{99}}{{100}} = \frac{1}{{100}}\\\\\left. \begin{array}{l} \Rightarrow \frac{1}{{{2^n}}} < \frac{1}{{100}}\\\\\frac{1}{{{2^7}}} = \frac{1}{{128}} < \frac{1}{{100}}\end{array} \right\} \Rightarrow n = 7\end{array}\)

\(\begin{array}{l}1\,,\,a\,,\,{a^2}\,,\,...\,,\,{a^{n - 1}}\\\\{S_n} = 1 + a + {a^2} + ... + {a^{n - 1}}\,\,\,(1)\\\\\left. \begin{array}{l}{a_1} = 1\\\\q = a\end{array} \right\}{S_n} = {a_1}\frac{{1 - {q^n}}}{{1 - q}} = 1 \times \frac{{1 - {a^n}}}{{1 - a}} = \frac{{{a^n} - 1}}{{a - 1}}\,\,\,(2)\\\\(1)\,,\,(2):\\\\1 + a + {a^2} + ... + {a^{n - 1}} = \frac{{{a^n} - 1}}{{a - 1}}\end{array}\)

\(\begin{array}{l}\frac{{{a^n} - 1}}{{a - 1}} = 1 + a + {a^2} + ... + {a^{n - 1}}\\\\{a^n} - 1 = (a - 1)({a^{n - 1}} + ... + {a^2} + a + 1)\end{array}\)