已知一个等比数列．．A2＝2．A5＝1／4．则A1A2＋A3A4＋．．．＋AN＋AN

A5 = A2 *q^(3)q^3 =1/8 q = 1/2A1=A2/q = 4An = A1q^(n-1) = 2^(3-n)设Bn = A(n-1)An,则：Bn = 2^(7-2n)=2^5 * (1/4)^(n-1)这是首项为2^5,公比为1/4的等比数列所以：A1A2＋A3A4＋．．．＋An-1An=128*[1-(1/4)^n]/3...