高数,数列向量综合题,数列{an},首项a1=-1,前n项和为sn,向量OB=a n-1向量OA-an向量OC,ABC共线,但直线不过原点,s20=?
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高数,数列向量综合题,数列{an},首项a1=-1,前n项和为sn,向量OB=a n-1向量OA-an向量OC,ABC共线,但直线不过原点,s20=?
高数,数列向量综合题,
数列{an},首项a1=-1,前n项和为sn,向量OB=a n-1向量OA-an向量OC,ABC共线,但直线不过原点,s20=?
高数,数列向量综合题,数列{an},首项a1=-1,前n项和为sn,向量OB=a n-1向量OA-an向量OC,ABC共线,但直线不过原点,s20=?
OB=a(n-1)OA-anOC
则：a(n-1)-an=1
不过这个需要证明才行,如需证明,
即：an-a(n-1)=-1
故：a2=a1-1=-2
a3=a2-1=-3
...
故：an=-n
故：S20=-(1+2+...+20)
=-(1+20)*20/2=-210
OB = a(n-1)OA-anOC (1)
ABC共线
let |AB|/|BC|=k
OB = (OA+kOC)/(1+k) (2)
from (1) and (2)
a(n-1) = 1/(1+k) (3) and
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OB = a(n-1)OA-anOC (1)
ABC共线
let |AB|/|BC|=k
OB = (OA+kOC)/(1+k) (2)
from (1) and (2)
a(n-1) = 1/(1+k) (3) and
-an= k/(1+k) (4)
from (3)
a1= 1/(1+k)
-1-k = 1
k = -2
from (4)
-an = -2/(1-2)
an = -2
ie
an = -1 ; n=1
= -2 ; n>=2
S20 = -1 -2(19)
= -39
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