如题所述
S9=a1(1-q^9)/(1-q)
S3=a1(1-q³)/(1-q)
相除则a1/(1-q)约分
所以=(1-q^9)/(1-q³)
=(1-q³)(1+q³+q^9)/(1-q³)
=1+q³+q^9
你的那个是S9/S6
S3=a1(1-q³)/(1-q)
相除则a1/(1-q)约分
所以=(1-q^9)/(1-q³)
=(1-q³)(1+q³+q^9)/(1-q³)
=1+q³+q^9
你的那个是S9/S6
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第1个回答 2012-07-30
S9=a1(1-q^9)/(1-q)
S3=a1(1-q^3)/(1-q)
S9/S3
=(1-q^9)/(1-q^3)
=(1-q^3)(1+q^3+q^6)/(1-q^3)
=1+q^3+q^6
S9/S6
=(1-q^9)/(1-q^6)
=(1-q^3)(1+q^3+q^6)/[(1-q^3)(1+q^3)]
=(1+q^3+q^6}/(1+q^3)追问
S3=a1(1-q^3)/(1-q)
S9/S3
=(1-q^9)/(1-q^3)
=(1-q^3)(1+q^3+q^6)/(1-q^3)
=1+q^3+q^6
S9/S6
=(1-q^9)/(1-q^6)
=(1-q^3)(1+q^3+q^6)/[(1-q^3)(1+q^3)]
=(1+q^3+q^6}/(1+q^3)追问
1-q^9为什么=(1-q^3)(1+q^3+q^6)
