用拆项法
Sn=(1+1/2)+(3+1/4)+...+[(2n-1)+1/2^n]
=[1+3+...+(2n-1)]+(1/2+1/4+...+1/2^n)
=n[1+(2n-1)]/2 +1/2(1-1/2^n)/(1-1/2)
=n^2+1/2-1/2^n
用拆项法
Sn=(1+1/2)+(3+1/4)+...+[(2n-1)+1/2^n]
=[1+3+...+(2n-1)]+(1/2+1/4+...+1/2^n)
=n[1+(2n-1)]/2 +1/2(1-1/2^n)/(1-1/2)
=n^2+1/2-1/2^n