求解1/(2^-5+√2)+ 1/(2^-4+√2)+ 1/(2^-3+√2)+……+1/(2^0+√2)+……+1/(2^6+√2)

1/(2^-5+√2)=1/(1/2^5+√2)
=2^5/(1+2^5√2)=(2^10√2-2^5)/(2^11-1)
1/(2^6+√2)=(2^6-√2)/(2^12-2)
=(2^6-√2)/[2(2^11-1)]
1/(2^-5+√2)+1/(2^6+√2)
=(2^6-√2)/[2(2^11-1)]+(2^10√2-2^5)/(2^11-1)
=(2^6-√2+2^11√2-2^6)/(2^11-1)
=√2/2
1/(2^-5+√2)+ 1/(2^-4+√2)+ 1/(2^-3+√2)+……+1/(2^0+√2)+……+1/(2^6+√2)
=6*(√2/2)=3√2

对任意非负整数k
1/[2^(-k)+√2]
=2^(k+1/2)/[2^(k+1)+√2] (分子分母同乘以2^(k+1/2))
=1/√2-1/[2^(k+1)+√2]
即1/[2^(-k)+√2]+1/[2^(k+1)+√2]=1/√2
故原式
=6/√2=3√2