Giải thích các bước giải:
Bài 1:
a,
\({u_1} = 5 - 3n = 5 - 3.\left[ {\left( {n - 1} \right) + 1} \right] = 2 - 3.\left( {n - 1} \right)\)
Suy ra \(\left( {{u_n}} \right)\) là CSC có \({u_1} = 2;\,\,\,\,d = - 3\)
b,
\({u_n} = \frac{{2n + 3}}{5} = \frac{3}{5} + \frac{2}{5}.n = \frac{3}{5} + \frac{2}{5}.\left[ {\left( {n - 1} \right) + 1} \right] = 1 + \frac{2}{5}\left( {n - 1} \right)\)
Suy ra \(\left( {{u_n}} \right)\) là CSC có \({u_1} = 1;\,\,\,d = \frac{2}{5}\)
c,
\(\left( {{u_n}} \right)\) không phải là CSC.
Bài 5c,
\(\begin{array}{l}
\left\{ \begin{array}{l}
{u_7} - {u_3} = 8\\
{u_2}.{u_7} = 75
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {{u_1} + 6d} \right) - \left( {{u_1} + 2d} \right) = 8\\
\left( {{u_1} + d} \right)\left( {{u_1} + 6d} \right) = 75
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
4d = 8\\
\left( {{u_1} + d} \right)\left( {{u_1} + 6d} \right) = 75
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
d = 2\\
\left( {{u_1} + 2} \right)\left( {{u_1} + 12} \right) = 75
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
d = 2\\
{u_1}^2 + 14{u_1} - 51 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
d = 2\\
\left[ \begin{array}{l}
{u_1} = 3\\
{u_1} = - 17
\end{array} \right.
\end{array} \right.\\
6,\\
a,\\
\left\{ \begin{array}{l}
{u_3} + {u_{13}} = 80\\
{S_{15}} = 600
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {{u_1} + 2d} \right) + \left( {{u_1} + 12d} \right) = 80\\
{u_1} + \left( {{u_1} + d} \right) + \left( {{u_1} + 2d} \right) + ...... + \left( {{u_1} + 14d} \right) = 600
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2{u_1} + 14d = 80\\
15{u_1} + \left( {1 + 2 + 3 + .... + 14} \right)d = 600
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} + 7d = 40\\
15{u_1} + \frac{{15.14}}{2}.d = 600
\end{array} \right.
\end{array}\)
Hệ trên có vô số nghiệm.
\(\begin{array}{l}
c,\\
{S_n} = 4{n^2} + 5n\\
\Leftrightarrow {u_1} + \left( {{u_1} + d} \right) + \left( {{u_1} + 2d} \right) + .... + \left[ {{u_1} + \left( {n - 1} \right)d} \right] = 4{n^2} + 5n\\
\Leftrightarrow n.{u_1} + \left( {1 + 2 + 3 + ..... + \left( {n - 1} \right)} \right)d = 4{n^2} + 5n\\
\Leftrightarrow n.{u_1} + \frac{{n\left( {n - 1} \right)}}{2}d = 4{n^2} + 5n\\
\Leftrightarrow {u_1} + \frac{{n - 1}}{2}d = 4n + 5\\
\Leftrightarrow \left( {{u_1} - \frac{1}{2}d} \right) + n.\frac{d}{2} = 4n + 5\\
\Leftrightarrow \left\{ \begin{array}{l}
{u_1} - \frac{1}{2}d = 5\\
\frac{d}{2} = 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{u_1} = 9\\
d = 8
\end{array} \right.
\end{array}\)
