
高二数学,求解
2个回答
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(1)解:设数列公比q,则
a1(1-q³)/(1-q)=3/2①
a1(1-q^6)/(1-q)=21/16②
②÷①,化简,有(1-q^6)/(1-q³)=7/8
1+q³=7/8
q=-1/2③
③ 代入①,得a1=2
an=2(-1/2)^(n-1)
(2)证明:2S10=S4+S7
2a1(1-q^10)/(1-q)=a1(1-q^4)/(1-q)+a1(1-q^7)/(1-q)
2a1(1-q^10)=a1(1-q^4)+a1(1-q^7)
2a1-2a1q^10=2a1-a1q^4-a1q^7
2a1q^10=a1q^4+a1q^7
等式两边都除以q^4
2a1q^6=a1+a1q^3
2a7=a1+a4
∴a1,a7,a4 也是等差数列
a1(1-q³)/(1-q)=3/2①
a1(1-q^6)/(1-q)=21/16②
②÷①,化简,有(1-q^6)/(1-q³)=7/8
1+q³=7/8
q=-1/2③
③ 代入①,得a1=2
an=2(-1/2)^(n-1)
(2)证明:2S10=S4+S7
2a1(1-q^10)/(1-q)=a1(1-q^4)/(1-q)+a1(1-q^7)/(1-q)
2a1(1-q^10)=a1(1-q^4)+a1(1-q^7)
2a1-2a1q^10=2a1-a1q^4-a1q^7
2a1q^10=a1q^4+a1q^7
等式两边都除以q^4
2a1q^6=a1+a1q^3
2a7=a1+a4
∴a1,a7,a4 也是等差数列
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