Question

已知等差数列{an}的首项a1≠0,公差d≠1,前n项和为sn,则s5n/s3n-s2n=

Answer 1

an = a1+(n-1)d

S(5n) = (2a1+(5n-1)d)(5n)/2 (1)

S(3n)-S(2n)=(2a1+(3n-1)d)(3n)/2 - (2a1+(2n-1)d)(2n)/2

=(n/2)[ 3(2a1+(3n-1)d) - 2(2a1+(2n-1)d) ]

=(n/2)[ 2a1+ (5n-1)d] (2)

(1)/(2)

S(5n)/(S(3n)-S(2n)) =5