Tìm cấp số cộng (Un), biết u1+u2+……+un=a và u1^2+u2^2+……+un^2=b^2 giải giúp em nha mọi người

Giải thích các bước giải:

\(\begin{array}{l}
1 + 2 + 3 + … + n = \frac{{n\left( {n + 1} \right)}}{2}\\
{1^2} + {2^2} + {3^2} + …. + {n^2} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\\
\left\{ \begin{array}{l}
{u_1} + {u_2} + {u_3} + …. + {u_n} = a\\
{u_1}^2 + {u_2}^2 + ….. + {u_n}^2 = {b^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{u_1} + \left( {{u_1} + d} \right) + \left( {{u_1} + 2d} \right) + …. + \left( {{u_1} + \left( {n – 1} \right)d} \right) = a\\
{u_1}^2 + {\left( {{u_1} + d} \right)^2} + {\left( {{u_1} + 2d} \right)^2}…. + {\left( {{u_1} + \left( {n – 1} \right)d} \right)^2} = {b^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
n{u_1} + \left( {1 + 2 + 3 + …. + \left( {n – 1} \right)} \right)d = a\\
n{u_1}^2 + \left( {2 + 4 + 6 + … + 2\left( {n – 1} \right)} \right){u_1}d + \left( {1 + {2^2} + … + {{\left( {n – 1} \right)}^2}} \right)d = {b^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
n{u_1} + \frac{{n\left( {n – 1} \right)}}{2}d = a\\
n{u_1}^2 + n\left( {n – 1} \right){u_1}d + \frac{{\left( {n – 1} \right).n\left( {2n – 1} \right)}}{6}{d^2} = {b^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{n^2}{u_1}^2 + n{u_1}.n\left( {n – 1} \right)d + \frac{{{n^2}{{\left( {n – 1} \right)}^2}}}{4}{d^2} = {a^2}\\
{n^2}{u_1}^2 + {n^2}\left( {n – 1} \right){u_1}d + \frac{{\left( {n – 1} \right).n\left( {2n – 1} \right)}}{6}{d^2} = {b^2}
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{n^2}{u_1}^2 + {n^2}\left( {n – 1} \right){u_1}d + \frac{{{n^2}{{\left( {n – 1} \right)}^2}}}{4}{d^2} = {a^2}\\
{n^2}{u_1}^2 + {n^2}\left( {n – 1} \right){u_1}d + \frac{{\left( {n – 1} \right).n\left( {2n – 1} \right)}}{6}{d^2} = {b^2}
\end{array} \right.\\
 \Leftrightarrow \frac{{{n^2}{{\left( {n – 1} \right)}^2}}}{4}{d^2} – \frac{{\left( {n – 1} \right)n\left( {2n – 1} \right)}}{6}{d^2} = {a^2} – {b^2}\\
 \Leftrightarrow \frac{{n\left( {n – 1} \right)}}{2}\left( {\frac{{n\left( {n – 1} \right)}}{2} – \frac{{2n – 1}}{3}} \right){d^2} = {a^2} – {b^2}\\
 \Rightarrow d = ….\\
 \Rightarrow {u_1} = ….
\end{array}\)