sinA+√3cosA=2
(sinA+√3cosA)²=4
sin²A+2√3sinAcosA+3cos²A=4
1/2-cos(2A)/2+√3sin(2A)+3/2+3cos(2A)/2=4
√3sin(2A)/2+cos(2A)/2=1
sin(2A)cos(π/6)+cos(2A)sin(π/6)=1
sin(2A+π/6)=1
A∈(0,π),则2A+π/6∈(0,13π/6)
sin(2A+π/6)=1,则2A+π/6=π/2,得A=π/6
由正弦定理,有a/sinA=b/sinB
2/sin(π/6)=b/sin(π/4)
得b=2sin(π/4)/sin(π/6)=2√2
sinC=sin[π-(A+B)]=sin(A+B)=sinAcosB+cosAsinB
=sin(π/6)cos(π/4)+cos(π/6)sin(π/4)=(√2+√6)/4
SΔABC=absinC/2=[2•2√2•(√2+√6)/4]/2=1+√3
该确定的三角形面积是1+√3
Sn=n²+2n
Sn-1=(n-1)²+2(n-1)
an=Sn-Sn-1=n²+2n-(n-1)²-2(n-1)=2n+1
通项公式an=2n+1
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