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P78961118 發表於 2018-6-10 08:14

等差和問題

請教各位老師,此題怎解,謝謝。
(無答案)
9.
假設\(a_1,a_2,a_3,\ldots,a_{11}\)為等差數列,公差為\(\sqrt{2}\),且總和為\(44\sqrt{5}\),試求
\( \displaystyle \frac{a_2^2}{a_1 \cdot a_3}+\frac{a_3^2}{a_2 \cdot a_4}+\frac{a_4^2}{a_3 \cdot a_5}+\frac{a_5^2}{a_4 \cdot a_6}+\frac{a_6^2}{a_5 \cdot a_7}+\frac{a_7^2}{a_6 \cdot a_8}+\frac{a_8^2}{a_7 \cdot a_9}+\frac{a_9^2}{a_8 \cdot a_{10}}+\frac{a_{10}^2}{a_9 \cdot a_{11}} \)之值。

CyberCat 發表於 2018-6-10 10:51

回復 1# P78961118 的帖子

試算了一下,參考看看:)
\(\displaystyle \frac{a_{k+1}^2}{a_{k}\cdot a_{k+2}}=\frac{(a_k+d)^2}{(a_k)(a_k+2d)}=\frac{a_k^2+2a_k d+d^2}{a_k^2+2a_k d}=1+\frac{d^2}{a_k^2+2a_k d}=1+\frac{d^2}{(a_k)(a_k+2d)}=1+d^2 \times \frac{1}{2}\left(\frac{1}{a_k}-\frac{1}{a_k+2d}\right)\)

\(\displaystyle \sum_{k=1}^9 \frac{a_{k+1}^2}{a_k \cdot a_{k+2}}=\sum_{k=1}^9 \left[1+d^2 \times \frac{1}{2}\left(\frac{1}{a_k}-\frac{1}{a_k+2d}\right) \right]=\sum_{k=1}^9 1+d^2 \times \frac{1}{2}\sum_{k=1}^9 \left(\frac{1}{a_k}-\frac{1}{a_k+2d}\right)\)

\(\displaystyle =9+(\sqrt{2})^2\times \frac{1}{2}\left[\left(\frac{1}{a_1}-\frac{1}{a_3}\right)+\left(\frac{1}{a_2}-\frac{1}{a_4}\right)+\ldots+\left(\frac{1}{a_9}-\frac{1}{a_{11}}\right)\right]
=9+\left(\frac{1}{a_1}+\frac{1}{a_2}-\frac{1}{a_{10}}-\frac{1}{a_{11}}\right)=9+\frac{\sqrt{2}}{2}\)

∵\(S_{11}=44\sqrt{5},d=\sqrt{2}\)

∴\(a_1=4\sqrt{5}-5\sqrt{2},a_2=4\sqrt{5}-4\sqrt{2},a_{10}=4\sqrt{5}+4\sqrt{2},a_{11}=4\sqrt{5}+5\sqrt{2}\)

\(\displaystyle \left(\frac{1}{a_1}+\frac{1}{a_2}-\frac{1}{a_{10}}-\frac{1}{a_{11}}\right)=\left(\frac{1}{a_1}-\frac{1}{a_{11}}\right)+\left(\frac{1}{a_2}-\frac{1}{a_{10}}\right)=\frac{a_{11}-a_1}{a_1\times a_{11}}+\frac{a_{10}-a_2}{a_2 \times a_{10}}\)

\(\displaystyle =\frac{(4\sqrt{5}+5\sqrt{2})-(4\sqrt{5}-5\sqrt{2})}{(4\sqrt{5}-5\sqrt{2})\times(4\sqrt{5}+5\sqrt{2})}+\frac{(4\sqrt{5}+4\sqrt{2})-(4\sqrt{5}-4\sqrt{2})}{(4\sqrt{5}-4\sqrt{2})\times(4\sqrt{5}+4\sqrt{2})}=\frac{10\sqrt{2}}{80-50}+\frac{8\sqrt{2}}{80-32}=\left(\frac{10}{30}+\frac{8}{48}\right)\sqrt{2}=\frac{\sqrt{2}}{2}\)

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