幂级数求和,:∑(n从1到正无穷) n*(n+2)*x^n

∑(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^(...