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當(dāng)q=2時(shí)a1 = a3/q² = 4/4 =1an = 1*q^(n-1) = 2^(n-1)當(dāng)q= -2/3時(shí)a4 = a3*q...（2）a1b1+a2b2+···+anbn=（2n-3)2^n+3,設(shè)數(shù)列{bn}的前n項(xiàng)和為Sn,求證：1/S1+1/S2+···+1/Sn小于等于2-1/n.第二問(wèn)能幫忙解答嗎2an= 2^(n-1)設(shè)cn = anbn 設(shè)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是等差數(shù)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< 1 + 1/(1*2) + 1/(2*3)+ ... + 1/(n-1)n= 1 + (1 -1/2) + (1/2 -1/3) +... + (1/n-1 - 1/n)= 2 -1/n