Question

設(shè)S_n,T_n分別為下列表格中的兩個(gè)等差數(shù)列{a_n},{b_n}的前n項(xiàng)和.

n	1	2	3	4	5	6	7	…
an	5	3	1	-1	-3	-5	-7	…
bn	-14	-10	-6	-2	2	6	10	…

(1)請(qǐng)求出S₁，S₂，S₄，S₅和T₂，T₃，T₅，T₆；

(2)根據(jù)上述結(jié)果，對(duì)于存在正整數(shù)k滿足a_k+a_k+1=0的等差數(shù)列{a_n}的前n項(xiàng)和S_n(n≤2k-1)的規(guī)律，猜想一個(gè)正確的結(jié)論，并加以證明.

Answer 1

解：(1)數(shù)列{a_n}中，a₁=5，公差d=-2.

∴S_n=na₁+d=5n+×(-2)=-n²+6n,

∴S₁=5,S₂=8,S₄=8,S₅=5.                                                        

數(shù)列{b_n}中,b₁=-14,d′=4.∴T_n=-14n+×4=2n²-16n,

∴T₂=-24,T₃=-30,T₅=-30,T₆=-24.                                                  

(2)若a_k+a_k+1=0,則S_k+n=S_k-n(或S_n=S_2k-n)(n≤2k-1,k＞n).

證明如下：∵{a_n}是等差數(shù)列，

∴S_k-n-S_k+n=(k-n)a₁+d-(k+n)a₁

=-2na₁+［(k-n)²-k+n-(k+n)²+k+n］

=-2na₁+(k²-2kn+n²+n-k²-2kn-n²+n)

=-2na₁+(-4kn+2n)=-2na₁+dn(1-2k).                                           

又∵a_k+a_k+1=0,∴a₁+(k-1)d+a₁+kd=0,d=,

∴S_k-n-S_k+n=-2na₁+(1-2k)=0，即S_k-n=S_k+n.