f(x)=2x+22xf(x)+f(1−x)=2x+22x+21−x+221−x=2x+22x+2+22x2=2x+22x+2=1 Now, k=1∑81f(82k)=f(821)+f(822)+……+f(8281)=f(821)+f(821)+……+f(1−822)+f(1−821) [f(821)+f(1−821)]+[f(822)+f(1−822)]+….40 cases +f(8241)(1+1+….+1)40 times +21/2+21/221/240+21=281