举报
tootoo222
你没说n从哪开始,我默认n从0到正无穷 注意到:(2n+1)x^2n=[x^(2n+1)]' ∑(2n+1)x^2n =∑[x^(2n+1)]' =[∑x^(2n+1)]' =[x∑x^2n]' =[x/(1-x²)]' =(1+x²)/(1-x²)² 中学方法,可以用错位相消法 设s(x)=1+3x²+5x^4+7x^6+.... 则:x²s(x)=x²+3x^4+5x^6+7x^8+.... 两式相减得:(1-x²)s(x)=1+2x²+2x^4+2x^6+... 则:(1-x²)s(x)=2/(1-x²)-1 因此s(x)=2/(1-x²)²-1/(1-x²)=(1+x²)/(1-x²)²