解：设S(x)=∑x^(2n+1)，∴S'(x)=∑(2n+1)x^(2n)。当丨x丨<1时，S(x)=∑x^(2n+1)=x/(1-x²)。∴∑(2n+1)x^(2n)=S'(x)=[x/(1-x²)]'=(1+x²)(1-x²)²。令x=1/√2，∴∑(2n+1)/2^n=(1+x²)(1-x²)²丨(x=1/√2)=6。供参考。
