Question

已知数列an是一个以1为首项,2/3为公差的等差数列,bn=(-1)^(n-1)*An*A(n+1),求数列bn的前n项和sn

Answer 1

由
s1
=
a1
=
[(a1
+
1)
/
2]^2
,
得
a1
=
1
,
所以
s2
=
1
+
a2
=
[(a2
+
1)
/
2]^2
,
得
a2
=
3
或
-1
，
因为数列{an}是等差数列，公差d>0,
所以
a2
=
3
,
所以
d
=
2
,
所以
an
=
2
n
-
1
,
所以
sn
=
n^2
,
所以
tn
=
-
1
+
2^2
-
3^2
+
4^2
-
5^2
+
6^2
-
…
若
n
为偶数，则
tn
=
(2+1)(2-1)+(4+3)(4-3)+(6+5)(6-5)+
…
+(n
+
n-1)[n
-
(n-1)]
=
1
+
2
+
3
+
4
+
…
+
n-1
+
n
=
n
(n
+
1)
/
2
,
若
n
为奇数，则
tn
=
(n
-
1)
n
/
2
-
n^2
=
-
n
(n
+
1)
/
2
,
综上，
tn
=
(-1)^n
*
n
*
(n
+
1)
/
2

Answer 2

本题考查的是数列重组后新数列的性质问题
当n=2k时,(相邻两项提公因式后,变成n/2个特殊数列公差为4/3)
Sn=b1+b2+...+b2k
=A1A2-A2A3+A3A4-A4A5+...+A(2k-1)A2k-A2kA(2k-1)
=A2(A1-A3)+A4(A3-A5)+...+A2k(A2k-1-A2k+1)
=-4/3(A2+A4+...A2k)
=-2n^2/9-2n/3
当n=2k-1时(去掉第一项后,相邻两项提公因式后重组成k-1项的
等差数列
)
Sn=b1+b2+b3+...+b(2k-1)
=A1A2-A2A3+A3A4-A4A5+...-A(2k-2)A(2k-1)+A(2k-1)A2k
=A1A2-A3(A2-A4)-A5(A4-A6)-...A(2k-1)[A(2k-2)-A(2k)]
=1*(5/3)+4/3[A3+A5+..A(2k-1)]
=2n^2/9+2n/3+7/9