问题
解答题
已知数列{an}的前n项和Sn是二项式(1+2x)2n(n∈N*)展开式中含x奇次幂的系数和. (1)求数列{an}的通项公式; (2)设f(n)=
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答案
(1)记(1+2x)2n=a0+a1x+a2x2+…+a2n-1x2n-1+a2nx2n
令x=1得:32n=a0+a1+a2+…+a2n-1+a2n
令x=-1得:1=a0-a1+a2-…-a2n-1+a2n
两式相减得:32n-1=2(a1+a3+…+a2n-1)
∴Sn=
(9n-1)(4分)1 2
当n≥2时,an=Sn-Sn-1=4×9n-1当n=1时,a1=S1=4,适合上式
∴an=4×9n-1(n∈N) (6分)
(2)f(n)=
=4 4×9n+12 1 9n+3
注意到f(n)+f(1-n)=
+1 9n+3
=1 91-n+3
+1 9n+3
=9n 9+3×9n
(8分)1 3
cn=f(0)+f(
)+f(1 n
)+…+f(2 n
),n n
可改写为cn=f(
)+f(n n
)+…+f(n-1 n
)+f(0)1 n
∴2cn=[f(0)+f(
)]+[f(n n
)+f(1 n
)]+…+[f(n-1 n
)+f(n-1 n
)]+[f(1 n
)+f(0)]n n
故cn=
,即f(0)+f(n+1 6
)+f(1 n
)+…+f(2 n
)=n n
(8分)n+1 6
∴
=1 cncn+1
=36×(36 (n+1)(n+2)
-1 n+1
)1 n+2
+1 c1c2
+…+1 c2c3 1 cncn+1
=36×[(
-1 2
)+(1 3
-1 3
)+…+(1 4
-1 n+1
) (12分)1 n+2
=36×(
-1 2
)]=18-1 n+2
(14分)36 n+2