886

已知数列 ${a_{n}}$ 满足 $a_{m + n} = \frac{1}{2} (a_{2 m} + a_{2 n}) - (m - n)^{2}, a_{0} = 11, a_{1} = 11$ , 求 $a_{2023}$ 的值.

解钱莘芪

注意到原式可变为 $\begin{array}{r} A_{m + n} - (m + n)^{2} = \frac{1}{2} [(a_{2 m} - (2 m)^{2}) + (a_{2 n} - (2 n)^{2})], \end{array}$
故 ${a_{n} - n^{2}}$ 是等差数列. 又 $a_{0} - 0 = 11, a_{1} - 1 = 10$ , 故公差为 $- 1$ . 这样 $\begin{array}{r} A_{2023} = 2023^{2} + (11 - 2023) = 4090517. \end{array}$