Question

bn=(n-2)(1/2)^(n-1)求证4Tn

Answer 1

Tn=(-1)(1/2)^0+0(1/2)^1+''''''''''''+(n-2)(1/2)^(n-1)

=2Tn-Tn

={(-1)*2+0(1/2)^0+1(1/2)^1+''''''''''+(n-2)(1/2)^(n-2)}-

{(-1)(1/2)^0+0(1/2)^1+1(1/2)^2+'''''''''''+(n-3)(1/2)^(n-2)+(n-2)(1/2)^(n-1)}

={2^0+2^(-1)+2^(-2)+''''''''+2^(2-n)}+(-2)-(n-2)(1/2)^(n-1)

求出Tn的表达式,问题就可以解决了