n(n?1) |
2 |
∴
Sn |
n |
(n?1) |
2 |
∴
Sn+1 |
n+1 |
Sn |
n |
d |
2 |
∴{
Sn |
n |
d |
2 |
∴
S12 |
12 |
S10 |
10 |
d |
2 |
又
S12 |
12 |
S10 |
10 |
∴d=2.
∵數(shù)列{an}為等差數(shù)列,a1=-2 012,
∴S2012=2012a1+
2012×(2012?1) |
2 |
=2012×(-2012)+
2012×(2012?1) |
2 |
=-2012.
故選B.
S12 |
12 |
S10 |
10 |
n(n?1) |
2 |
Sn |
n |
(n?1) |
2 |
Sn+1 |
n+1 |
Sn |
n |
d |
2 |
Sn |
n |
d |
2 |
S12 |
12 |
S10 |
10 |
d |
2 |
S12 |
12 |
S10 |
10 |
2012×(2012?1) |
2 |
2012×(2012?1) |
2 |