摘要：设f(k)是满足不等式log2x+log2(3·2k-1-x)≥2k-1(k∈N*)的自然数x的个数. (1)求f(k)的表达式, (2)记Sn=f(1)+f(2)+-+f(n).Pn=n2+n-1.当n≤5时试比较Sn与Pn的大小. 解:(1)由不等式log2x+log2(3·2k-1-x)≥2k-1.得x(3·2k-1-x)≥22k-1.解之得2k-1≤x≤2k.故f(k)=2k-2k-1+1=2k-1+1. (2)∵Sn=f(1)+f(2)+-+f(n)=1+2+22+23+-+2n-1+n=2n+n-1. ∴Sn-Pn=2n+n-1-(n2+n-1)=2n-n2. 又n≤5.可计算得S1＞P1.S2=P2.S3＜P3.S4=P4.S5＞P5.

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