记
　　f(x) = ∑(n=1~inf.)[(n-1)x^(2n-2)]/3^n
　= (1/3)∑(n=1~inf.)n[(x^2)/3]^(n-1) - (1/3)∑(n=1~inf.)[(x^2)/3]^(n-1)
　=(1/3)f1(x) - (1/3)f2(x),
而
　　f2(x) = ∑(n=1~inf.)[(x^2)/3]^(n-1)
　= 1/[1 - (x^2)/3],|(x^2)/3| < 1,
又
　　　　∫[0,x] f1(t)dt
= ∑(n=1~inf.)n∫[0,x]{[(t^2)/3]^(n-1)}dt
= ∑(n=1~inf.)[(t^2)/3]^n
　= 1/[1 - (x^2)/3] - 1,|(x^2)/3| < 1,
可得
　　f1(x) = (d/dx)∫[0,x] f1(t)dt
　　　= (d/dx) {1/[1 - (x^2)/3] - 1}
　　　　= ……,|(x^2)/3| < 1,
由此可得结果,……