Question

等比数列{an}前n项和与积分别为S和T,数列{1/an}的前n项和为P,求证:T^2=(S/P)^n

Answer 1

证明：当公比q=1时,S=na1,T=a1^n,p=n/a1

(S/P)^n=(na1/(n/a1))^n=(a1^2)^n=a1^(2n)=(a1^n)^2=T^2.

当公比q≠1时,

an=a1q^(n-1)(n 为自然数）

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

T=a1*a2*...*an=a1*(a1q)*(a1q^2)*.*(a1q^(n-1))

=a1^n*q^(1+2+3+...+n-1)

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

P=1/a1+1/a2+...+1/an=1/a1+1/(a1q)+...+1/(a1q^(n-1)

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

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

=q(q^n-1)/(a1q^n(q-1))

(S/P)^n={[a1(1-q^n)/(1-q)]/[q(q^n-1)/(a1q^n(q-1))]}^n

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

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

=(（a1^n*q^(n(n-1)/2) )^2

=T^2