1、在三角形ABC中,角ABC的各边分别为abc 已知 b cosC=(2a-c)cosB

等我明天 (1) 已知 b cosC=(2a-c)cosB b cosC+c cosB=2acosBsinB * cosC+sinc* cosB=2sinA*cosB(正弦定理） sin(B+C)=sinA=2sinA*cosBcosB=1/2 B=60 [a²+c²-b²]/2ac=cosB=1/2 b²=ac a²+c²=2ac(a-c)²=0a=cB=60ΔABC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^nSn=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√3/3sin75=sin(45+30)=[√6+√2]/4 c/sinC=a/sinA a=[3+√3]/2SΔABC=1/2acsinB=3[3+√3]/8（4)an=19+(n-1)*(-2)=-2n+21Sn=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+20nTn=[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-3bn=n^2+2n1/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)]=ok6 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)+1an+1=2[(a(n-1)+1][an+1]/[(a(n-1)+1]=2等比公比2首项2an+1=2*2^(n-1)=2^nan=2^n-1sn=(2^(n+1)-1)-nok