等我明天
(1) 已知 b cosC=(2a-c)cosB
b cosC+c cosB=2acosB
sinB * cosC+sinc* cosB=2sinA*cosB(正弦定理）
sin(B+C)=sinA=2sinA*cosB
cosB=1/2
B=60
[a²+c²-b²]/2ac=cosB=1/2 b²=ac
a²+c²=2ac
(a-c)²=0
a=c
B=60
ΔABC是等边三角形
（2)已知a1+2a2=0 s4-s2=1/8
a1+2a1q=0 a1(1+2q)=0 q=-1/2
[a1(1-q^3)/(1-q)]-]a1(1-q^2)/(1-q)]=1/8 a1.q^2=1/8 a1=1/2
an=-1/2（1/2）^(n-1)=-1/2^n
Sn=a1(1-q^n)/(1-q)]=1/2^n-1
(3）已知sin2C=sinAcosB+cosAsinB
2sinCcosC =sinAcosB+cosAsinB=sin(A+B) = sinC cosC=1/2 C=60
cosC的平方-1/sin(A+B)cos（A+B）=cos60²-1/sin120/cos120
=1/2^2-1/√3/2/(-1/2)
=1/4+4√3/3
sin75=sin(45+30)=[√6+√2]/4 c/sinC=a/sinA a=[3+√3]/2
SΔABC=1/2acsinB=3[3+√3]/8
（4)an=19+(n-1)*(-2)=-2n+21
Sn=na1+1/2n(n-1)d=-n^2+20n
已知bn-an=3^(n-1)
bn=an+3^(n-1 )=3^(n-1 )-n^2+20n
Tn=[3^(n-1 )-1]/2-1/6n(n+1)(2n+1)+10(1+n)n=ok
(5) S6=6(a1+a6）/2=3（a1+a6)=60 a1+a6=20 2a1+5d=20
已知 a6^2=a1*a21 a1=2.5d 带入上式得
d=2 a1=5
故an=2n+3 Sn=(2n+3+5)*n/2=n^2+4n
已知b(n+1) - bn=an=2n+3
bn-b(n-1)=2(n-1）+3
b(n-1)-b(n-2)=2(n-3)+3
:
b2-b1=2+3
等式两边相加得：
bn-b1=2*(1+n-1)(n-1)/2+3(n-1)=n^2+2n-3
bn=n^2+2n
1/bn=1/(n^2+2n) =1/2[1/n-1/(n+2)]
T n=1/2[1-1/3+1/2-1/4+1/3-1/5+……-1/n+1/(n+1)-1/(n+2)]=1/2[1+1/2-1/n-1/(n+2)]=ok
6 已知 2a sinA =(2b+c)sinB+(b+2c)sinC
2a^2=2b^2+2c^2+2bc
a^2-b^2-c^2=bc cosA=-1/2 ∠A=120 sinB=sinC=sin30=1/2 三角形ABC的形状等腰三角形
7. a n+1=2 s n =2（an-an-1)
an=2a(n-1)+1
an+1=2[(a(n-1)+1]
[an+1]/[(a(n-1)+1]=2等比公比2首项2
an+1=2*2^(n-1)=2^n
an=2^n-1
sn=(2^(n+1)-1)-n
ok
