已知数列{an}是各项都是整数的等比数列,a3=4,{an}的前三项和等于7.（1）求数列{an}的通项公式

a1+a2 = S3-a3 = 7-4 =3a2 = a1*qa3 = a1*q²a3/(a1+a2) = q²/(1+q) = 4/3得3q² -4q -4 =0(q-2)(3q+2)=0q=2 或 q= -2/3当q=2时a1 = a3/q² = 4/4 =1an = 1*q^(n-1) = 2^(n-1)当q= -2/3时a4 = a3*q...（2）a1b1+a2b2+···+anbn=（2n-3)2^n+3,设数列{bn}的前n项和为Sn,求证：1/S1+1/S2+···+1/Sn小于等于2-1/n.第二问能帮忙解答吗2an= 2^(n-1)设cn = anbn 设a1b1+a2b2+···+anbn = Tncn = anbn = bn * 2^(n-1) cn = Tn - Tn-1 = (2n-3)2^n+3 - [(2(n-1)-3)2^(n-1) +3]= 2(2n-3)2^(n-1) - (2n-5)2^(n-1)= (2n -1) 2^(n-1)所以cn = anbn = bn * 2^(n-1)= (2n -1) 2^(n-1)bn = (2n-1)bn是等差数列，公差=2，首相=1Sn = (1+(2n-1))/2 * n = n²1/S1+1/S2+···+1/Sn = 1/1 + 1/4 + 1/9 +... + 1/n²