已知Sn是等比數(shù)列an的前n項(xiàng)和,S4S10S7成等差數(shù)列,若a1=1,求數(shù)列an^3的前n項(xiàng)的積
S4=a1(1-q^4)/(1-q)
S10=a1(1-q^10)/(1-q)
S7=a1(1-q^7)/(1-q)
,S4S10S7成等差數(shù)列,
S4-S10=S10-S7
a1(1-q^4)/(1-q)-a1(1-q^10)/(1-q)=a1(1-q^10)/(1-q)-a1(1-q^7)/(1-q)
2q^6-q^3-1=0
(2q^3+1)(q^3-1)=0
q1=1
q2=(-1/2)^(1/3)
a1=1,q=1
等比數(shù)列an=1
前n項(xiàng)的積=1
a1=1,q=(-1/2)^(1/3)
an=a1*q^(n-1)=-1/2^[(n-1)/3]
an^3=(-1/2^[(n-1)/3])^3=-1/2^(n-1)
前n項(xiàng)的積=(-1)^(n-1)/2^[n(n+1)/2] (n>1)
a2=我覺得 那個(gè)2 的指數(shù) 應(yīng)該是n(n-1)/2,按照等差數(shù)列求和來看，首相是0，末項(xiàng)是n-1 對(duì)么？？還是謝謝你yes,前n項(xiàng)的積=(-1)^(n-1)/2^[n(n-1)/2] (n>1)