Question

若数列{a_n}满足a₁＝2且a_n＋a_n－1＝2ⁿ＋2^n－1，S_n为数列{a_n}的前n项和，则log₂(S_{2 012}＋2)＝________.

Answer 1

2013

因为a₁＋a₂＝2²＋2，a₃＋a₄＝2⁴＋2³，a₅＋a₆＝2⁶＋2⁵，…，
所以S_{2 012}＝a₁＋a₂＋a₃＋a₄＋…＋a_{2 011}＋a_{2 012}＝2¹＋2²＋2³＋2⁴＋…＋2^{2 011}＋2^{2 012}＝＝2^{2 013}－2.
故log₂(S_{2 012}＋2)＝log₂2^{2 013}＝2 013.