Question

设f(k)是满足不等式log₂x+log₂(3·2^k-1-x)≥2k-1(k∈N^*)的自然数x的个数.

(1)求f(k)的表达式；

(2)记S_n=f(1)+f(2)+…+f(n),P_n=n²+n-1,当n≤5时试比较S_n与P_n的大小.

Answer 1

解:(1)由不等式log₂x+log₂(3·2^k-1-x)≥2k-1,得x(3·2^k-1-x)≥2^2k-1,

    解之,得2^k-1≤x≤2^k,

    故f(k)=2^k-2^k-1+1=2^k-1+1.

    (2)∵S_n=f(1)+f(2)+…+f(n)=1+2+2²+2³+…+2^n-1+n=2ⁿ+n-1,

    ∴S_n-P_n=2ⁿ+n-1-(n²+n-1)=2ⁿ-n².