给出下列四个结论: ①已知△ABC中,三边a,b,c满足(a+b+c)(a+b-c)=3ab,则∠C等于120°. ②若等差数列an的前n项和为Sn,则三点(10,
③等差数列an中,若S10=30,S20=100,则S30=210. ④设f(x)=
其中,结论正确的是 ______.(将所有正确结论的序号都写上) |
①由(a+b+c)(a+b-c)=3ab,得到(a+b)2-c2=3ab,化简得:a2+b2-c2=ab,
则cosC=
=a2+b2-c2 2ab
=ab 2ab
,根据C∈(0,180°),得到∠C=60°,所以此选项错误;1 2
②因为
=S10 10
=a1+10a1+
d10×9 2 10
d,同理9 2
=a1+S100 100
d,99 2
=a1+S110 110
d,109 2
则
=
-S100 100 S10 10 100-10
=(a1+
d)-(a1+99 2
d) 9 2 90
=d 2
=
-S110 110 S100 100 110-100
=(a1+
d)-(a1+109 2
d) 99 2 10
,d 2
所以三点(10,
),(100,S10 10
),(110,S100 100
)共线.此选项正确;S110 110
③根据等差数列的性质可知,S10,S20-S10,S30-S20成等差数列,
得到:2(S20-S10)=S10+(S30-S20),将S10=30,S20=100,
代入得:2(100-30)=30+(S30-100),解得:S30=210.此选项正确;
④因为f(x)+f(1-x)=
+1 2x+ 2 1 21-x+ 2
=
+1 2x+ 2
=2x 2 +
•2x2
+2
(2x+2
)2 2x 2+
•2x2
=
=
+2x2 1 +
•2x2
=
+2x2
(2
+2x)2
,2 2
则f(-8)+f(-7)+…+f(0)+…+f(8)+f(9)=
×9=2 2
.此选项正确.9 2 2
所以,正确的结论序号有:②③④.
故答案为:②③④