171819     三角形面积为4分之3倍根号320     求bn的前n项和公式2122  求Tn

17、（1）∵a = 2bsinA ∴ a/sinA = 2b = b/sinB ∴ sinB = 1/2 ∴ B = 30°
（2）∵a = 3√3；c=5；
∴b² = a² + c² -2accosB = 27 + 25 -30√3 × √3/2 = 7 ∴ b = √7
18、由题意知道：x²+ax+b = (x+5)(x-3) = x²+2x-15 ∴ a =2,b=-15
∴2x²+x-15 < 0 的解集为：{x | -3< x < 5/2}
19、（1）∵b²+c²-a² = bc ∴ cosA = (b²+c²-a²)/2bc = 1/2 ∴ A = 60°
（2）∵b=1 ∴b边上的高 h = 2S÷b = 3√3/2 ；∴c = h/sinA = 3
20、（1）a6 = a3+(6-3)d ∴ d = 2 ∴ an = -6+2(n-3) = 2n-12
(2) ∵ b1 = -8 ；b2 = a1+a2+a3 = 2(1+2+3)-12×3 = -24 ∴ q = 3
{bn}的Sn = 4(1-3^n)
21、略.
22、（1）Sn = n²+2n ； S(n-1) = (n-1)²+2n-2 = n²-1
∴an = Sn - S(n-1) = 2n+1
（2）Tn = 1/2[1/3 - 1/(2n+3)] = n/(6n+9)

17. （1）由a/sinA=b/sinB
得：2bsinA/sinA=b/sinB
2b=b/sinB
b/2b=sinB
...

17.b/sinB=a/sinA=2b,sinB=1/2,又因B为锐角,B=30度
b^2=a^2+c^2-2accosB=27+25-45=7,b=根号7

n>1时，an=Sn-Sn-1=n2+2n-(n-1)2-2(n-1)=2n+1;
n=1时，a1=S1=3,符合公式an=2n+1,g
故an=2n+1;
Tn=1/a1a2+1/a2a3+⋯1/(anan+1)=1/(3*5)+1/(5*7)+⋯1/((2n+1)(2n+3))=1/2*(1/3-1/5+1/5-1/7+⋯-1/(2n...

大题你一个都不会？？？