请各位知道做的朋友帮帮忙……请用高二所学的知识帮忙解答,①求和：(a-1)+(a的二次方-2)+(a的三次方-3)+(a
3个回答
(a-1)+(a^2-2)_(a^3-3)+...+(a^n-n)
=(a+a^2+a^3+...+a^n)-(1+2+3+...+n)
=a(1-a^n)/(1-a) -(1+n)n/2
a^2-c^2=b^2-bc
a^2=b^2+c^2-2bccosA
b^2-2bccosA=b^2-bc
cosA=1/2
A=60
sinA=2sinBsinC
sinA=sin(180-B-C)=sin(B+C)=sinBcosC+cosBsinC
sinBcosC-sinBsinC=sinBsinC-cosBsinC
sinB(cosC-sinC)=sinC(sinB-cosB)
(cosC-sinC)/sinC=(sinB-cosB)/sinB
cotC+cotB=2
cotA=cot(180-B-C)=-cot(B+C)=-1/tan(B+C)=(tanBtanC-1)/(tanB+tanC)
=(1-cotBcotC)/(cotC+cotB)
cotA=√3 1-cotBcotC=2√3 cotBcotC=1-2√3
cotBcotC0,cotC90或C>90
钝角三角形
3
S1=a1 S2=a1+a1q S3=a1+a1q+a1q^2
S2-S1=S3-S2
2S2=S1+S3
2+2q=1+1+q+q^2
q^2-q=0
q=1
如果S1,mS2,S3成等差数列
2*(mS2)=S1+S3
2m+2mq=2+q+q^2
q^2-(2m-1)q-(2m-2)=0
(q-(2m-1)/2)^2=(2m-2)+(2m-1)^2/4
q=[(2m-1)+√(4m^2+4m-7)]/2
