補上出處
設數列\( a_n=\root 3 \of{n^2+2n+1}+\root 3 \of{n^2-1}+\root 3 \of{n^2-2n+1} \),則\( \displaystyle \frac{1}{a_1}+\frac{1}{a_3}+\frac{1}{a_5}+...+\frac{1}{a_{2001}} \)
(90高中數學能力競賽 宜花東區試題,h ttp://www.math.nuk.edu.tw/senpengeu/HighSchool/2002_Taiwan_High_Ilan_02.pdf 連結已失效)
對每個\( n \in N \),設\( a_n=\root 3 \of{n^2+2n+1}+\root 3 \of{n^2-1}+\root 3 \of{n^2-2n+1} \),則\( \displaystyle \sum_{n=1}^{500} \frac{1}{a_{2n-1}} \)的值是?
(1991上海高中數學競賽)
設數列\( a_n=\root 3 \of{n^2+2n+1}+\root 3 \of{n^2-1}+\root 3 \of{n^2-2n+1} \),則\( \displaystyle \sum_{n=1}^{1000} \frac{1}{a_n}= \)?
(94蘭陽女中)
設對所有的正整數n,\( a_n=\root 3 \of{n^2+2n+1}+\root 3 \of{n^2-1}+\root 3 \of{n^2-2n+1} \),\( \displaystyle \frac{1}{a_1}+\frac{1}{a_3}+\frac{1}{a_5}+...+\frac{1}{a_{997}}+\frac{1}{a_{999}}= \)?
(A)1 (B)3 (C)5 (D)7
(95基隆市國中聯招)
\( g(n)=\root 3 \of{n^2+2n+1}+\root 3 \of{n^2-1}+\root 3 \of{n^2-2n+1} \).
\( \displaystyle \frac{1}{g(1)}+\frac{1}{g(3)}+...+\frac{1}{g(999999)} \)
https://artofproblemsolving.com/community/c6h333174
113.5.12補充
若\(n\)為正整數,且\(\displaystyle a_n=\root 3\of{(n+1)^2}+\root 3\of{n^2-1}+\root 3\of {(n-1)^2}\),試求\(\displaystyle \frac{1}{a_1}+\frac{1}{a_3}+\frac{1}{a_5}+\ldots+\frac{1}{a_{4095}}=\)
。
(113內湖高工,
https://math.pro/db/thread-3866-1-1.html)