Question

数学证明

求1-4的证明过程
求1-5的答案，不需要过程

Answer 1

令Sn= k=(0,n) 和(k-1)/2^k= (0-1)/2^0 + (1-1)/2^1+...+ (n-2)/2^(n-1)+ (n-1)/2^n
2Sn= (0-1)/2^(0-1)+(1-1)/2^(1-1)+ (2-1)/2^(2-1)+...+(n-1)/2^(n-1)
相减得到Sn=(0-1)/2^(0-1)+1/2^0+1/2^1+...+1/2^(n-1)-(n-1)/2^n
=-2+(1-(1/2)^n)/(1-1/2)-(n-1)/2^n
=-2(1/2)^n-(n-1)/2^n
n->无穷Sn->0

1-5
Sn=(2+1)x^2+(2*2+1)x^4 +...+(2n+1)x^(2n)
x^2Sn = (2+1)x^4+ (2*2+1)x^6+...+(2n-1)x^(2n)+ (2n+1)x^(2n+2)
相减
(1-x^2)Sn=(2+1)x^2+2x^4+2x^6+...+2x^(2n)- (2n+1)x^(2n+2)
Sn=[ (2+1)x^2+2x^4+2x^6+...+2x^(2n)- (2n+1)x^(2n+2) ]/ (1-x^2)
=[x^2+2x^2(1-(x^2)^n)/(1-x^2)- (2n+1)x^(2n+2) ]/(1-x^2)
若|x|>=1,n->无穷，发散
若|x|<1,n->无穷
Sn->[x^2+2x^2/(1-x^2)]/(1-x^2)=(3x^2-x^4)/(1-x^2)^2