Sn为等差数列{an}的前n项和，若a2nan＝4n−12n−1，则S2nSn=．

Question

S_n为等差数列{a_n}的前n项和，若(a2n)/(an)＝(4n−1)/(2n−1)，则(S2n)/(Sn)=______．

Answer 1

解析：答  由(a2n)/(an)＝(4n−1)/(2n−1)，
即 (an+nd)/(an)＝(4n−1)/(2n−1)，得an＝(2n−1)/(2)d，a1＝(d)/(2)．
Sn＝(n(a1+an))/(2)＝(n2d)/(2)，S2n＝((2n)2d)/(2)＝4Sn．
故(S2n)/(Sn)=4．
故答案为4．