如果数列
1
,
1
+
2
,
1
+
2
+
4
,
⋯
 
,
1
+
2
+
2
2
+
2
3
+
⋯
+
2
n
−
1
,
⋯
1,1+2,1+2+4,\cdots,1+2+2^2+2^3+\cdots+2^{n-1},\cdots
1 , 1 + 2 , 1 + 2 + 4 , ⋯ , 1 + 2 + 2 2 + 2 3 + ⋯ + 2 n − 1 , ⋯
n
n
n
S
n
>
1020
S_n>1020
S n > 1 0 2 0
n
n
n
    
\;\;
A
.
7
A. 7
A . 7
B
.
8
B. 8
B . 8
C
.
9
C. 9
C . 9
D
.
10
D. 10
D . 1 0
[解析]
S
n
=
n
⋅
2
0
+
(
n
−
1
)
⋅
2
1
+
(
n
−
2
)
⋅
2
2
+
⋯
+
1
⋅
2
n
−
1
⋯
(
1
)
S_n=n\cdot2^0+(n-1)\cdot2^1+(n-2)\cdot2^2+\cdots+1\cdot2^{n-1}\quad\cdots\quad(1)
S n = n ⋅ 2 0 + ( n − 1 ) ⋅ 2 1 + ( n − 2 ) ⋅ 2 2 + ⋯ + 1 ⋅ 2 n − 1 ⋯ ( 1 )
2
S
n
=
n
⋅
2
1
+
(
n
−
1
)
⋅
2
2
+
(
n
−
2
)
⋅
2
3
+
⋯
+
1
⋅
2
n
⋯
(
2
)
2S_n=n\cdot2^1+(n-1)\cdot2^2+(n-2)\cdot2^3+\cdots+1\cdot2^{n}\quad\cdots\quad(2)
2 S n = n ⋅ 2 1 + ( n − 1 ) ⋅ 2 2 + ( n − 2 ) ⋅ 2 3 + ⋯ + 1 ⋅ 2 n ⋯ ( 2 )
S
n
=
−
n
+
2
1
+
2
2
+
⋯
+
2
n
−
1
+
2
n
=
2
n
+
1
−
n
−
2
S_n=-n+2^1+2^2+\cdots+2^{n-1}+2^n=2^{n+1}-n-2
S n = − n + 2 1 + 2 2 + ⋯ + 2 n − 1 + 2 n = 2 n + 1 − n − 2
S
9
=
1013
  
,
  
S
10
=
2035
S_9=1013\;,\;S_{10}=2035
S 9 = 1 0 1 3 , S 1 0 = 2 0 3 5
[法二]
a
n
−
a
n
−
1
=
2
n
−
1
a_n-a_{n-1}=2^{n-1}
a n − a n − 1 = 2 n − 1
a
2
−
a
1
=
2
a_2-a_1=2
a 2 − a 1 = 2
a
3
−
a
2
=
2
2
a_3-a_2=2^2
a 3 − a 2 = 2 2
a
4
−
a
3
=
2
3
a_4-a_3=2^3
a 4 − a 3 = 2 3
⋯
\cdots
⋯
a
n
−
a
n
−
1
=
2
n
−
1
a_n-a_{n-1}=2^{n-1}
a n − a n − 1 = 2 n − 1
a
n
=
1
+
2
+
2
2
+
⋯
+
2
n
−
1
=
2
n
−
1
a_n=1+2+2^2+\cdots+2^{n-1}=2^n-1
a n = 1 + 2 + 2 2 + ⋯ + 2 n − 1 = 2 n − 1
S
n
=
2
n
+
1
−
n
−
2.
S_n=2^{n+1}-n-2.
S n = 2 n + 1 − n − 2 .