
这个函数怎么做,谢谢!
1个回答
2018-07-15
展开全部
consider
1/(1-x) = 1+x+x^2+....
两边取导数
1/(1-x)^2 = 1+2x+3x^2+....
两边乘以 x^3
x^3/(1-x)^2 = x^3+2x^4+3x^5+....
两边取导数
[3x^2(1-x)^2 + 2x^3. (1-x) ]/(1-x)^4 = (1)(3)x^2+(2)(4)x^3+....
x^2.(1-x).[ 3(1-x)+2x ]/(1-x)^4 = (1)(3)x^2+(2)(4)x^3+....
x^2.(3-x)/(1-x)^3 = (1)(3)x^2+(2)(4)x^3+....
两边除以 x
x(3-x)/(1-x)^3 = (1)(3)x+(2)(4)x^2+....
=>
∑(n: 1->∞ ) n(n+2)x^n
=x(3-x)/(1-x)^3
1/(1-x) = 1+x+x^2+....
两边取导数
1/(1-x)^2 = 1+2x+3x^2+....
两边乘以 x^3
x^3/(1-x)^2 = x^3+2x^4+3x^5+....
两边取导数
[3x^2(1-x)^2 + 2x^3. (1-x) ]/(1-x)^4 = (1)(3)x^2+(2)(4)x^3+....
x^2.(1-x).[ 3(1-x)+2x ]/(1-x)^4 = (1)(3)x^2+(2)(4)x^3+....
x^2.(3-x)/(1-x)^3 = (1)(3)x^2+(2)(4)x^3+....
两边除以 x
x(3-x)/(1-x)^3 = (1)(3)x+(2)(4)x^2+....
=>
∑(n: 1->∞ ) n(n+2)x^n
=x(3-x)/(1-x)^3
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