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.求证（1）S/R=a1an(2)P²R²=S的n次方麻烦老师能细写过程,和说明解题道理,

Question

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.求证（1）S/R=a1an
(2)P²R²=S的n次方 麻烦老师能细写过程,和说明解题道理,

Answer 1

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