f=sum((2*q*cos(2* i*x)*(-1)^(i)*(-1)^((2*i-1)))/(i*Pi),i=1.3.5...35)

I want to write this series but getting error

the result is

2*q*cos(2*x)/Pi-2*cos(6*x)*q/(3*Pi)+2*q*cos(10*x)/(5*Pi)-2*q*cos(14*x)/(7*Pi)+2*q*cos(18*x)/(9*Pi)-2*q*cos(22*x)/(11*Pi)+2*q*cos(26*x)/(13*Pi)-2*q*cos(30*x)/(15*Pi)+2*q*cos(34*x)/(17*Pi)-2*q*cos(38*x)/(19*Pi)+2*q*cos(42*x)/(21*Pi)-2*q*cos(46*x)/(23*Pi)+2*q*cos(50*x)/(25*Pi)-2*q*cos(54*x)/(27*Pi)+2*q*cos(58*x)/(29*Pi)-2*q*cos(62*x)/(31*Pi)+2*q*cos(66*x)/(33*Pi)-2*q*cos(70*x)/(35*Pi)

can anybody help 

eq2 := -4*A[2]*cos(2*x)-16*A[4]*cos(4*x)-36*A[6]*cos(6*x)-64*A[8]*cos(8*x)+a*A[0]+cos(8*x)*a*A[8]+cos(6*x)*a*A[6]+cos(4*x)*a*A[4]+2*cos(2*x)*q*A[0]+cos(2*x)*a*A[2]+q*A[8]*cos(6*x)+q*A[8]*cos(10*x)+q*A[6]*cos(4*x)+q*A[6]*cos(8*x)+q*A[4]*cos(2*x)+q*A[4]*cos(6*x)+q*A[2]*cos(4*x)+q*A[2]

How to extract the co-fficent that do not have cos in it like

a*A[0]+q*A[2]

N := 4;
print(`output redirected...`); # input placeholder
4
y := sum(A[2*n].cos(2.*n.x), n = 0 .. N);

eq1 := diff(y, `$`(x, 2))+(a+2*q*cos(2*x))*y

eq2 := map(combine, eq1, trig)

for i from 0 to 4 do eq4[i] := coeff(eq2, cos(2*n*x)) end do

From these I want to extract the co-ffficients of cos(0x),cos(2x),cos(4x)..

and form a simultaneous linear equation containg A0,A2,A4

The solution is 

aA0+qA2=0

2q*A0+(a-4)*A2+q*A4=0

Can anybody tell me how to do it