※ 引述《semmy214 (黃小六)》之銘言：
: https://imgur.com/a/OKaSA
: 其他題求過程
數列題
令 b_n = √a_n, 則 b_1 = 1, b_n = 2b_{n+1} + b_n * b_{n+1}
這可以寫成 (b_n - 2b_{n+1}) / (b_n * b_{n+1}) = 1
或是 1/b_{n+1} - 2/b_n = 1
所以再令 c_n = 1/b_n, 則 c_1 = 1, c_{n+1} - 2c_n = 1
這就容易解了, 易解得 c_n = 2^n - 1
故 b_n = 1/(2^n - 1), a_n = 1/(2^n - 1)^2
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提一下思路, 首先把數列開根號這應該很容易想到
第三行的關鍵變形可以在嘗試求出 b_n 的前幾項 1, 1/3, 1/7, 1/15 之後觀察出來
要的是想辦法做出 1/b_n 的形式所以進行那樣的變形
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