{U1+U2+U3+U4=30 AND U1^2+U2^2+U3^2+U4^2= 340 GIÚP MÌNH VSSSSS Cấp số nhân tìm u1, q

Ta có: \(\left\{ \begin{array}{l}{u_1} + {u_2} + {u_3} + {u_4} = 30\\u_1^2 + u_2^2 + u_3^2 + u_4^2 = 340\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{u_1} + q{u_1} + {q^2}{u_1} + {q^3}{u_1} = 30\\u_1^2 + {q^2}u_1^2 + {q^4}u_1^2 + {q^6}u_1^2 = 340\end{array} \right.\)

\( \Leftrightarrow \left\{ \begin{array}{l}{u_1}\left( {1 + q + {q^2} + {q^3}} \right) = 30\,\,\left( 1 \right)\\u_1^2\left( {1 + {q^2} + {q^4} + {q^6}} \right) = 340\,\,\left( 2 \right)\end{array} \right.\)

Nếu \(q = 1\) thì \(\left\{ \begin{array}{l}4{u_1} = 30\\4u_1^2 = 340\end{array} \right. \Rightarrow {u_1} = \dfrac{{34}}{3}\)

Nếu \(q =  – 1\) thì \({u_1}.0 = 30\left( {vo\,li} \right)\) nên \(q =  – 1\) không thỏa mãn.

Nếu \(q \ne  \pm 1\) thì:

Ta có: \(1 + q + {q^2} + {q^3} = \dfrac{{{q^4} – 1}}{{q – 1}};1 + {q^2} + {q^4} + {q^6} = \dfrac{{{q^8} – 1}}{{{q^2} – 1}}\)

Lấy (2) chia (1) vế với vế ta được

\({u_1}.\dfrac{{1 + {q^2} + {q^4} + {q^6}}}{{1 + q + {q^2} + {q^3}}} = \dfrac{{340}}{{30}} \Leftrightarrow {u_1}.\dfrac{{{q^8} – 1}}{{{q^2} – 1}}:\dfrac{{{q^4} – 1}}{{q – 1}} = \dfrac{{340}}{{30}}\)

\( \Leftrightarrow {u_1}.\dfrac{{{q^4} + 1}}{{q + 1}} = \dfrac{{34}}{3} \Leftrightarrow {u_1} = \dfrac{{34\left( {q + 1} \right)}}{{3\left( {{q^4} + 1} \right)}}\)

Thay vào phương trình (1) được:

\(\dfrac{{34\left( {q + 1} \right)}}{{{q^4} + 1}}.\dfrac{{{q^4} – 1}}{{q – 1}} = 30 \Leftrightarrow \dfrac{{34\left( {{q^5} + {q^4} – q – 1} \right)}}{{{q^5} – {q^4} + q – 1}} = 30\) \( \Leftrightarrow 34{q^5} + 34{q^4} – 34q – 34 = 30{q^5} – 30{q^4} + 30q – 30\)

\( \Leftrightarrow 4{q^5} + 64{q^4} – 64q – 4 = 0 \Leftrightarrow {q^5} + 16{q^4} – 16q – 1 = 0\)

\( \Leftrightarrow \left( {q – 1} \right)\left( {{q^4} + 17{q^3} – 17{q^2} – 17q + 1} \right) = 0\)

\(\begin{array}{l} \Leftrightarrow {q^4} + 17{q^3} – 17{q^2} – 17q + 1 = 0\\ \Leftrightarrow {q^2} + 17q – 17 – \dfrac{{17}}{q} + \dfrac{1}{{{q^2}}} = 0\\ \Leftrightarrow \left( {{q^2} + \dfrac{1}{{{q^2}}}} \right) + 17\left( {q – \dfrac{1}{q}} \right) – 17 = 0\\ \Leftrightarrow {\left( {q – \dfrac{1}{q}} \right)^2} + 2 + 17\left( {q – \dfrac{1}{q}} \right) – 17 = 0\\ \Leftrightarrow {\left( {q – \dfrac{1}{q}} \right)^2} + 17\left( {q – \dfrac{1}{q}} \right) – 15 = 0\\ \Leftrightarrow \left[ \begin{array}{l}q – \dfrac{1}{q} = \dfrac{{ – 17 + \sqrt {349} }}{2}\\q – \dfrac{1}{q} = \dfrac{{ – 17 – \sqrt {349} }}{2}\end{array} \right….\end{array}\)