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#741ca0d1-298e-4f46-9d89-be909a05965a中等解答题数列求通项方法数列

14.已知数列{an}\{a_{n}\}的前nn项和为SnS_{n}a1=1a_{1}=1nSn+1=(n+2)SnnS_{n+1}=(n+2)S_{n}. (1)求{an}\{a_{n}\}的通项公式; (2)bkb_{k}为满足kan2kk\leqslant a_{n}\leqslant 2^{k}ana_{n}的个数,求使b1+b2++bk>2023b_{1}+b_{2}+\ldots +b_{k}>2023成立的最小正整数kk的值.