说明:此题只能在│x│∫∑(n+1)x^ndx=C+x+x²+x³+.+x^n+. (C是积分常数) (两端同时积分)
=C+x/(1-x)
==>∑(n+1)x^ndx=1/(1-x)² (两端同时求导)
∴∑(2n-1)x^n=2/(1-x)²-3/(1-x)
=(3x-1)/(1-x)².
说明:此题只能在│x│∫∑(n+1)x^ndx=C+x+x²+x³+.+x^n+. (C是积分常数) (两端同时积分)
=C+x/(1-x)
==>∑(n+1)x^ndx=1/(1-x)² (两端同时求导)
∴∑(2n-1)x^n=2/(1-x)²-3/(1-x)
=(3x-1)/(1-x)².