内容正文:
第二课时
等差数列前n项和的性质及应用(习题课)
等差数列的前n项和性质的应用
[例1] (1)等差数列前n项的和为30,前2n项的和为100,则它的前3n项的和为
( )
A.130 B.170 C.210 D.260
(2)等差数列{an}共有2n+1项,所有的奇数项之和为132,所有的偶数项之和为120,则n等于________;
(3)已知{an},{bn}均为等差数列,其前n项和分别为Sn,Tn,且eq \f(Sn,Tn)=eq \f(2n+2,n+3),则eq \f(a5,b5)=________.
[解析] (1)利用等差数列的性质:
Sn,S2n-Sn,S3n-S2n成等差数列.
所以Sn+(S3n-S2n)=2(S2n-Sn),
即30+(S3n-100)=2(100-30),解得S3n=210.
(2)因为等差数列共有2n+1项,所以S奇-S偶=an+1=eq \f(S2n+1,2n+1),即132-120=eq \f(132+120,2n+1),解得n=10.
(3)由等差数列的性质,知
eq \f(a5,b5)=eq \f(\f(a1+a9,2),\f(b1+b9,2))=eq \f(\f(a1+a9,2)×9,\f(b1+b9,2)×9)=eq \f(S9,T9)=eq \f(2×9+2,9+3)=eq \f(5,3).
[答案] (1)C (2)10 (3)eq \f(5,3)
等差数列的前n项和常用的性质
(1)等差数列的依次k项之和,Sk,S2k-Sk,S3k-S2k,…组成公差为k2d的等差数列;
(2)数列{an}是等差数列⇔Sn=an2+bn(a,b为常数)⇔数列eq \b\lc\{\rc\}(\a\vs4\al\co1(\f(Sn,n)))为等差数列;
(3)若S奇表示奇数项的和,S偶表示偶数项的和,公差为d:
①当项数为偶数2n时,S偶-S奇=nd,eq \f(S奇,S偶)=eq \f(an,an+1);
②当项数为奇数2n-1时,S奇-S偶=an,eq \f(S奇,S偶)=eq \f(n,n-1).
[跟踪训练]
1.设等差数列{an}的前n项和为Sn,若S4=8,S8=20,则a11+a12+a13+a14=
( )
A.18
B.17
C.16
D.15
解析:设{a