Question

在等差数列{an}中,证明(a1+a2+..+a2n-1)/(2n-1)=an 类比上述性质

Answer 1

等差数列设an=a1+k(n-1)

(a1+a2+..+a2n-1)/(2n-1)

=((2n-1)*a1+k(0+1+2+3+...+2n-2))/(2n-1)

=((2n-1)*a1+k*(2n-2)*(2n-1)/2)/(2n-1)

=a1+k*(n-1)

=an

类比

等比数列有

(a1*a2*...*a2n-1)开2n-1次方根=an

证明：

等比数列设an=a1*k^(n-1)

(a1*a2*...*a2n-1)

=(a1*k^0*a1*k^1*.*a1*k^(2n-2))

=(a1^(2n-1)*k^(0+1+.+2n-2))

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

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

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

所以(a1*a2*...*a2n-1)开2n-1次方根

=(a1*k^(n-1))^(2n-1)开2n-1次方根

=a1*k^(n-1)

=an