If ar is the coefficient of xr in the expansion of 1-2x+3x2nthen∑r=02nr ar=

We have1-2x+3x2n=a0+a1x+a2x2+…+a2nx2n⇒ddx1-2x+3x2n=ddxa0+a1x+…+a2nx2n⇒n1-2x+3x2n-1(-2+6x)=a1+a2·2x+a3·3x2+…a2n2n·x2n-1 putting x=1, we get n2n-1(4)=a1+a2·2+a3·3+…a2n·2n⇒∑r=12nr.ar=n·2n+1