设数列{an}中是等比数列,且S=a1+a2+---an,R=1/a1+1/a2+---1/an,P=a1a2---an

1个回答

  • P²R²(R^n?)=S的n次方

    s=a1(1-q^n)/(1-q)

    r=1/a1+1/a2+.+1/an

    r/q=1/a2+1/a3+.+1/a(n+1)

    r-r/q=1/a1-1/a(n+1)

    (1-1/q)r=1/a1-1/a(n+1)

    (q-1)r/q=1/a1-1/a(n+1)

    r=[1/a1-1/a(n+1)]q/(q-1)

    r=[a(n+1)/a1a(n+1)-a1/a1a(n+1)]q/(q-1)

    r=[a(n+1)-a1]q/[a1a(n+1)(q-1)]

    1/r=[a1a(n+1)(q-1)]/[a(n+1)-a1]q

    s/r

    =a1(1-q^n)/(1-q)*[a1a(n+1)(q-1)]/[a(n+1)-a1]q

    =a1(1-q^n)*[a1a(n+1)]/[a1-a(n+1)]q

    =[a1-a(n+1)]*[a1a(n+1)]/[a1-a(n+1)]q

    =a1a(n+1)/q

    =a1anq/q

    =a1an

    p=a1a2.an

    =(a1)^n*q^(1+2+.+n-1)

    =(a1)^n*q^[n(n-1)/2]

    p²=(a1)^2n*q^n(n-1)

    =[(a1)^2q^(n-1)]^n

    =(a1*an)^n

    s/r=a1an

    r=s/(a1an)

    p²r^n

    =(a1*an)^n*s^n/(a1an)^n

    =s^n