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