Question

∑(-i)^(n-1)(2n-1)/2^nz^(2n-1)求收敛半径

Answer 1

这个数列通项是一个等差2n-1乘一个等比1/(2^(n-1))的形式%D%A所以正式错位相减法的典型题型%Dª(n)=(2n-1)/(2^(n-1))%D%AS(n)=1*1+3*(1/2)+5*(1/4)+...+(2n-1)*1/(2^(n-1))%D%A根据错位相减法,上式两边同时乘以等比的公比1/2,然后两式相减：%D%A(1/2)S(n)=1*(1/2)+3*(1/4)+...+(2n-3)*1/(2^(n-1))+(2n-1)*1/(2^n)%D%A注意上面对齐的方式,我已经给你错开位了,直接对应相减就行了：%D%A(1/2)S(n)=1*1+（2*(1/2)+2*(1/4）+2*(1/8）+...+2*1/(2^(n-1)）-(2n-1)*1/(2^n)%D%A整理得:(中间那部分就是等比了,你应该会求,用公式就行)%D%A(1/2)S(n)=1+2(1/2+1/4+...+1/(2^(n-1))-(2n-1)*1/(2^n)%D%A =1+2*(1-(1/2)^n-1)-(2n-1)*1/(2^n)%D%A =3+1/2^n-3/2^(n-1)