已知数列an的通项公式an=(4n-5)*(1/2)^(n-1),试猜测an的最大值并通过研究数列an的单调性证明结论,

f(x) = (4x-5)。(1/2)^(x-1)f'(x) = (1/2)^(x-1) 。[ 4 - (4x-5) ln2] =04 - (4x-5) ln2=0x= (4+5ln2)/(4ln2) =2。69an=(4n-5)*(1/2)^(n-1)a2= 3(1/2) = 3/2a3 = 7(1/2)^2 = 7/4max an = a3 = 7/4an is increasing 1≤n≤3an is decreasing n ≥3an = (4n-5)。(1/2)^(n-1)= 8(n。(1/2)^n) - 5(1/2)^(n-1)Sn = 8[∑(i:1->n) i。(1/2)^i] - 10[ 1-(1/2)^n]letS = 1。(1/2)+2(1/2)^2+。。。+n(1/2)^n (1)(1/2)S =1。(1/2)^2+2(1/2)^3+。。。+n(1/2)^(n+1) (2)(1)-(2)(1/2)S = [(1/2) + (1/2)^2+(1/2)^3+。。。+(1/2)^n] - n。(1/2)^(n+1)= 1-(1/2)^n - n。(1/2)^(n+1)S = 2[1-(1/2)^n - n。(1/2)^(n+1)]Sn = 8[∑(i:1->n) i。(1/2)^i] - 10[ 1-(1/2)^n]=8S -10[ 1-(1/2)^n]=16[1-(1/2)^n - n。(1/2)^(n+1)] - 10[ 1-(1/2)^n]= 6 -2(4n+3)。(1/2)^n