数列{a n }是公差为正数的等差数列,a 2 、a 5 且是方程x 2 -12x+27=0的两根,数列{b n }的前n项和为T n

2025-05-12 13:39:35
推荐回答(1个)
回答(1):

(1)∵等差数列{a n }的公差d>0,a 2 、a 5 且是方程x 2 -12x+27=0的两根,
∴a 2 =3,a 5 =9.
∴d=
9-3
5-2
=2,
∴a n =a 2 +(n-2)d=3+2(n-2)=2n-1;
又数列{b n }中,T n =1-
1
2
b n ,①
∴T n+1 =1-
1
2
b n+1 ,②
②-①得:
b n+1
b n
=
1
3
,又T 1 =1-
1
2
b 1 =b 1
∴b 1 =
2
3

∴数列{b n }是以
2
3
为首项,
1
3
为公比的等比数列,
∴b n =
2
3
? (
1
3
)
n-1

综上所述,a n =2n-1,b n =
2
3
? (
1
3
)
n-1

(2)∵c n =a n ?b n =(2n-1)?
2
3
? (
1
3
)
n-1

∴S n =a 1 b 1 +a 2 b 2 +…+a n b n
=1×
2
3
+3×
2
3
×
1
3
+…+(2n-1)×
2
3
× (
1
3
)
n-1
,③
1
3
S n =
2
3
×
1
3
+3×
2
3
× (
1
3
)
2
+…+(2n-3)×
2
3
× (
1
3
)
n-1
+(2n-1)×
2
3
× (
1
3
)
n
,④
∴③-④得:
2
3
S n =
2
3
+
4
3
[
1
3
+ (
1
3
)
2
+ (
1
3
)
3
+…+ (
1
3
)
n-1
]-(2n-1)×
2
3
× (
1
3
)
n

S n =1+2[
1
3
+ (
1
3
)
2
+ (
1
3
)
3
+…+ (
1
3
)
n-1
]-(2n-1)× (
1
3
)
n

=1+2×
1
3
[1- (
1
3
)
n-1
]
1-
1
3
-(2n-1)× (
1
3
)
n

=2-
2n+2
3
× (
1
3
)
n-1

=2-(2n+2)× (
1
3
)
n