等比数列an中,S2=7 S6=91,则S4=s2=a1(1-q^2)/(1-q)=7 ① s6=a1(1-q^6)/(1-q)=91 ② ②/①,得q^4+q^2-12=0,q^2=3代入①得,(2)a1/(1-q)=7/(1-3)=-7/2 s4=a1(1-q4)/(1-q)=(-7/2)*(1-3^2)=28(2)中为什么②/①=q^4+q^2-12=0
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![等比数列an中,S2=7 S6=91,则S4=s2=a1(1-q^2)/(1-q)=7 ① s6=a1(1-q^6)/(1-q)=91 ② ②/①,得q^4+q^2-12=0,q^2=3代入①得,(2)a1/(1-q)=7/(1-3)=-7/2 s4=a1(1-q4)/(1-q)=(-7/2)*(1-3^2)=28(2)中为什么②/①=q^4+q^2-12=0](/uploads/image/z/11184416-8-6.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97an%E4%B8%AD%2CS2%3D7+S6%3D91%2C%E5%88%99S4%3Ds2%3Da1%281-q%5E2%29%2F%281-q%29%3D7+%E2%91%A0+s6%3Da1%281-q%5E6%29%2F%281-q%29%3D91+%E2%91%A1+%E2%91%A1%2F%E2%91%A0%2C%E5%BE%97q%5E4%2Bq%5E2-12%3D0%2Cq%5E2%3D3%E4%BB%A3%E5%85%A5%E2%91%A0%E5%BE%97%2C%282%29a1%2F%281-q%29%3D7%2F%281-3%29%3D-7%2F2+s4%3Da1%281-q4%29%2F%281-q%29%3D%28-7%2F2%29%2A%281-3%5E2%29%3D28%282%29%E4%B8%AD%E4%B8%BA%E4%BB%80%E4%B9%88%E2%91%A1%2F%E2%91%A0%3Dq%5E4%2Bq%5E2-12%3D0)
等比数列an中,S2=7 S6=91,则S4=s2=a1(1-q^2)/(1-q)=7 ① s6=a1(1-q^6)/(1-q)=91 ② ②/①,得q^4+q^2-12=0,q^2=3代入①得,(2)a1/(1-q)=7/(1-3)=-7/2 s4=a1(1-q4)/(1-q)=(-7/2)*(1-3^2)=28(2)中为什么②/①=q^4+q^2-12=0
等比数列an中,S2=7 S6=91,则S4=
s2=a1(1-q^2)/(1-q)=7 ①
s6=a1(1-q^6)/(1-q)=91 ②
②/①,得q^4+q^2-12=0,q^2=3代入①得,(2)
a1/(1-q)=7/(1-3)=-7/2
s4=a1(1-q4)/(1-q)=(-7/2)*(1-3^2)=28
(2)中为什么②/①=q^4+q^2-12=0
等比数列an中,S2=7 S6=91,则S4=s2=a1(1-q^2)/(1-q)=7 ① s6=a1(1-q^6)/(1-q)=91 ② ②/①,得q^4+q^2-12=0,q^2=3代入①得,(2)a1/(1-q)=7/(1-3)=-7/2 s4=a1(1-q4)/(1-q)=(-7/2)*(1-3^2)=28(2)中为什么②/①=q^4+q^2-12=0
解由②/①得
[a1(1-q^2)/(1-q)]/[a1(1-q^6)/(1-q)]=7/91
即(1-q^2)/(1-q^6)=1/13
即(1-q^6)=13(1-q^2)
即(1-(q^2)^3)=13(1-q^2)
即(1-q^2)(1+q^2+q^4)=13(1-q^2)
即(1+q^2+q^4)=13
即q^4+q^2-12=0.
②/①得(1-q⁶)/(1-q²)=(1-q²)(1+q²+q⁴)/(1-q²)=1+q²+q⁴=13,故得q⁴+q²-12=0.
其中,分子1-q⁶=1-(q²)³=(1-q²)(1+q²+q⁴),是按立方差公式分解因式。