17,18题 过程要详细一点,谢谢啦
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(17)
(1)S=(1/2)(c²-a²-b²+2ab)
=(1/2)(2ab-2abcosC)
=ab(1-cosC)=(1/2)absinC
2-2cosC=sinC
5cos²C-8cosC+3=0
(5cosC-3)(cosC-1)=0
cosC=1(舍去)
cosC=3/5;
(2)
2sinAcosC=sinB=sin(A+C)=sinAcosC+cosAsinC
cos(A-C)=0
A=C
b=csinB/sinC=2csinCcosC/sinC=2ccosC=12/5
(18)
(1)
2a4=3a3-a2 【估计一a2是-a2的意思,这里用-a2了】
2a1q³=3a1q²-a1q
2q²-3q+1=0
得:q=1/2 ,q=1(舍去)
S6=a1(1-q^5)/(1-q)
=31a1/16=63/32 【估计是62/32……】
a1=63/62
an=a1q^(n-1)=63/62×2^(n-1)
(2)为计算方便(1)中的a1=1,an=2^(1-n)
bn=nan=n2^(1-n)
Tn=1+2/2+3/4+4/8+……n/2^(n-1)
Tn/2=1/2+2/4+3/8+……n/2^(n-2)
两式相减
Tn/2=1+1/2+1/4+……1/2^(n-1)-n/2^(n-2)
=2(1-1/2^n)-n/2^(n-2)
Tn=4(1-1/2^n)-2n/2^(n-2)
=4-(2n+1)/2^(n-2)
如果一定要那怪头怪脑的a1=63/62,就勇敢的a1Tn=63Tn/62算下去
(1)S=(1/2)(c²-a²-b²+2ab)
=(1/2)(2ab-2abcosC)
=ab(1-cosC)=(1/2)absinC
2-2cosC=sinC
5cos²C-8cosC+3=0
(5cosC-3)(cosC-1)=0
cosC=1(舍去)
cosC=3/5;
(2)
2sinAcosC=sinB=sin(A+C)=sinAcosC+cosAsinC
cos(A-C)=0
A=C
b=csinB/sinC=2csinCcosC/sinC=2ccosC=12/5
(18)
(1)
2a4=3a3-a2 【估计一a2是-a2的意思,这里用-a2了】
2a1q³=3a1q²-a1q
2q²-3q+1=0
得:q=1/2 ,q=1(舍去)
S6=a1(1-q^5)/(1-q)
=31a1/16=63/32 【估计是62/32……】
a1=63/62
an=a1q^(n-1)=63/62×2^(n-1)
(2)为计算方便(1)中的a1=1,an=2^(1-n)
bn=nan=n2^(1-n)
Tn=1+2/2+3/4+4/8+……n/2^(n-1)
Tn/2=1/2+2/4+3/8+……n/2^(n-2)
两式相减
Tn/2=1+1/2+1/4+……1/2^(n-1)-n/2^(n-2)
=2(1-1/2^n)-n/2^(n-2)
Tn=4(1-1/2^n)-2n/2^(n-2)
=4-(2n+1)/2^(n-2)
如果一定要那怪头怪脑的a1=63/62,就勇敢的a1Tn=63Tn/62算下去
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