等差数列{an}满足a1+a(2n-1)=2n,f(n)=S2n-Sn,请速回答
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a(n)=a+(n-1)d,

2n = a(1)+a(2n-1)=a + a+(2n-2)d=2[a + (n-1)d]=2a(n),

a(n)=n.

S(n)=n(n+1)/2,

S(2n)=2n(2n+1)/2=n(2n+1).

f(n)=S(2n)-S(n)=n(2n+1)-n(n+1)/2=n[4n+2-n-1]/2=n[3n+1]/2.

f(n+1)=(n+1)(3n+4)/2,

f(n+1)-f(n)=(n+1)(3n+4)/2 - n(3n+1)/2 = [3n^2+7n+4-3n^2-n]/2

=[6n+4]/2=3n+2>3*1+2=5>0,[n>=1]

2^x - 2^(-x) - t < 12f(n) = 12*n(3n+1)/2 = 6n(3n+1),

t > 2^x - 2^(-x) - 6n(3n+1)

2^x - 2^(-x) - 6n(3n+1)

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