幂级数求和，求高人指导！: ∑(n从1到正无穷) n*(n+2)*x^n
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∑(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
=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
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