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2024届高考冲刺卷(二)2数学试卷答案

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23-129B·

分析由a_n+2+a_ncosnπ=1，当n=2k-1时，k∈Z，a_2k+1-a_2k-1=1，可得数列{a_2k-1}是首项为1，公差为1的等差数列．当n=2k时，k∈Z，a_2k+2+a_2k=1．可得S₁₂₀．又a₆₁=31，即可得出．

解答解：由a_n+2+a_ncosnπ=1，当n=2k-1时，k∈Z，a_2k+1-a_2k-1=1，∴数列{a_2k-1}是首项为1，公差为1的等差数列．
∴a₁+a₃+…+a₁₁₉=$\frac{（1+60）×60}{2}$=1830．
当n=2k时，k∈Z，a_2k+2+a_2k=1．
∴a₂+a₄+…+a₁₂₀=（a₂+a₄）+（a₆+a₈）+…+（a₁₁₈+a₁₂₀）=30．
∴S₁₂₀=1830+30=1860．
又a₆₁=a_2×30+1=1+30=31，
∴$\frac{{S}_{120}}{{a}_{61}}$=$\frac{1860}{31}$=60．
故选：C．

点评本题考查了等差数列的定义及其前n项和公式、“分组求和”，考查了推理能力与计算能力，属于中档题．