回覆 19# jperica05 的帖子
填充2
設\( \left\{{a}_{n}\right\},\left\{{b}_{n}\right\}\)的公差分別為d , D
\( {S}_{n} = \frac{\left[2{a}_{1} + (n - 1)d\right] \cdot n}{2},{T}_{n} = \frac{\left[2{b}_{1} + (n - 1)D\right] \cdot n}{2}\)
\( \frac{{S}_{n}}{{T}_{n}} = \frac{2{a}_{1} + (n - 1)d}{2{b}_{1} + (n - 1)D} = \frac{2n + 1}{4n - 2}\)
\( \frac{{S}_{1}}{{T}_{1}} = \frac{{a}_{1}}{{b}_{1}} = \frac{3}{2},\mathrm{故可設}{a}_{1} = 3k,{b}_{1} = 2k,\mathrm{其中}k \neq 0\)
\( \frac{{S}_{2}}{{T}_{2}} = \frac{6k + d}{4k + D} = \frac{5}{6}\Rightarrow 5D - 6d = 16k\)
\( \frac{{S}_{3}}{{T}_{3}} = \frac{6k + 2d}{4k + 2D} = \frac{7}{10}\Rightarrow 14D - 20d = 32k\)
32k = 2( 5D - 6d ) = 14D - 20d , 可得 D = 2d
16k = 5D - 6d = 4d , d = 4k
\(\mathrm{求值式} = \frac{{a}_{1} + 9d}{{b}_{1} + 2D + {b}_{1} + 17D} + \frac{{a}_{1} + 10d}{{b}_{1} + 5D + {b}_{1} + 14D} = \frac{2{a}_{1} + 19d}{2{b}_{1} + 19D} = \frac{6k + 19d}{4k + 38d} = \frac{6k + 19 \cdot 4k}{4k + 38 \cdot 4k} = \frac{82k}{156k} = \frac{41}{78} \)
