等比數(shù)列{Ａn}的各項(xiàng)均為正數(shù)，其前n項(xiàng)中，數(shù)值最大一項(xiàng)是５４，若該數(shù)列前n項(xiàng)之和為Sn，且Sn＝８０，S2n＝６５６０．（１）求S100（２）求Ａn 請(qǐng)寫出過程，謝謝．

熱心網(wǎng)友

對(duì)于等比數(shù)列,有An=A1*q^(n-1),Sn=A1*(q^n-1)/(q-1),所以Sn=A1*(q^n-1)/(q-1)=80,S2n=A1*(q^(2n)-1)/(q-1)=6560,兩式相除,化簡(jiǎn),得到(q^n)^2 - 82*q^n + 81 = 0,求解得q^n=1或q^n=81,顯然本題中后者為正解.現(xiàn)根據(jù)常識(shí)判斷只有q=3,n=4才會(huì)q^n=81.因此數(shù)列遞增.于是An=A1*q^(n-1)=A1*q^n/q=A1*81/3=54,則A1=2故:An = 2 * 3^(n-1); S100 = 2 * (3^100 - 1)/(3 - 1) = 3^100 - 1