回覆 19# jperica05 的帖子
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已知\(S_n\)、\(T_n\)分別為等差數列\(\{\;a_n \}\;\)、\(\{\;b_n \}\;\)的首\(n\)項的和，且\(\displaystyle \frac{S_n}{T_n}=\frac{2n+1}{4n-2}(n=1,2,\ldots)\)。求\(\displaystyle \frac{a_{10}}{b_3+b_{18}}+\frac{a_{11}}{b_6+b_{15}}\)之值。
[解答]
設\( \left\{{a}_{n}\right\},\left\{{b}_{n}\right\}\)的公差分別為\(d,D\)
\(\displaystyle {S}_{n} = \frac{\left[2{a}_{1} + (n - 1)d\right] \cdot n}{2},{T}_{n} = \frac{\left[2{b}_{1} + (n - 1)D\right] \cdot n}{2}\)
\(\displaystyle \frac{{S}_{n}}{{T}_{n}} = \frac{2{a}_{1} + (n - 1)d}{2{b}_{1} + (n - 1)D} = \frac{2n + 1}{4n - 2}\)
\(\displaystyle \frac{{S}_{1}}{{T}_{1}} = \frac{{a}_{1}}{{b}_{1}} = \frac{3}{2},\mathrm{故可設}{a}_{1} = 3k,{b}_{1} = 2k,\mathrm{其中}k \neq 0\)
\(\displaystyle \frac{{S}_{2}}{{T}_{2}} = \frac{6k + d}{4k + D} = \frac{5}{6}\Rightarrow 5D - 6d = 16k\)
\(\displaystyle \frac{{S}_{3}}{{T}_{3}} = \frac{6k + 2d}{4k + 2D} = \frac{7}{10}\Rightarrow 14D - 20d = 32k\)
\(32k = 2( 5D - 6d ) = 14D - 20d\), 可得\(D = 2d\)
\(16k = 5D - 6d = 4d\), \(d = 4k\)
\(\displaystyle \mathrm{求值式} = \frac{{a}_{1} + 9d}{{b}_{1} + 2D + {b}_{1} + 17D} + \frac{{a}_{1} + 10d}{{b}_{1} + 5D + {b}_{1} + 14D} = \frac{2{a}_{1} + 19d}{2{b}_{1} + 19D} = \frac{6k + 19d}{4k + 38d} = \frac{6k + 19 \cdot 4k}{4k + 38 \cdot 4k} = \frac{82k}{156k} = \frac{41}{78} \)
