问题描述: 幂级数求和,:∑(n从1到正无穷) n*(n+2)*x^n 1个回答 分类:数学 2014-10-25 问题解答: 我来补答 ∑(n从1到正无穷) n(n+2)x^n=x∑(n从1到正无穷) n(n+2)x^(n-1)=x∑(n从1到正无穷)[(n+2)x^n]′=x[∑(n从1到正无穷)(n+2)x^n]′∑(n从1到正无穷)(n+2)x^n=1/x[∑(n从1到正无穷)(n+2)x^(n+1)]=1/x∑(n从1到正无穷)[x^(n+2)]′=1/x[∑(n从1到正无穷)x^(n+2)]′=1/x[x³/(1-x)]′=x(3-2x)/(1-x)²原式=x[x(3-2x)/(1-x)²]′=x(3-x)/(1-x)³ 展开全文阅读