Question

求解这三道题，在线等，急

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Answer 1

17
f(x)=sin(2x+π/3)+sin(2x-π/3)+2(cosx)^2
=1/2sin2x+√3/2cosx+1/2sin2x-√3/2cosx+(1+cos2x)
=sin2x+cos2x+1
=√2(√2/2sin2x+√2/2cos2x)+1
=√2sin(2x+π/4)+1
T=2π/2=π
(2)x在[-π/4,π/4]
2x+π/4在[-π/4,3π/4]
sin(2x+π/4)在[-π/4,3π/4]上值域为:[-√2/2,1]
f(x)=√2sin(2x+π/4)+1在[-π/4,3π/4]上值域为:[0,1+√2]
所以f(x)=√2sin(2x+π/4)+1在[-π/4,π/4]上最大值=1+√2
19
Sn=n^2+2n ,S(n-1)=(n-1)^2+2(n-1)
an=Sn-S(n-1)=n^2+2n-(n-1)^2-2(n-1)
an=2n+1
(2)bn=(an-5)/2^n=(n-2)*(1/2)^(n-1)
bn=(n-2)*(1/2)^(n-1)
Tn=2*(1/2)^4+4*(1/2)^5...+(n-2)*(1/2)^2n
1/2Tn=2(1/2)^5+4*(1/2)^6...+(n-2)*(1/2)^(2n+1)
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