分组求和(1+1)+(2+1/2)+(3+1/2的二次方+4+1/2的三次方+.n+1/2的n-1次方

an =n
Sn = a1+a2+...+an
= n(n+1)/2
bn = (1/2)^(n-1)
Tn = b1+b2+...+bn
= 2( 1- (1/2)^n )
(1+1)+(2+1/2)+(3+1/2^2+(4+1/2^3)+...+(n+1/2^(n-1) )
=(1+2+...+n) + [1+1/2 +1/2^n+...+1/2^(n-1) ]
= Sn +Tn
=n(n+1)/2 + 2( 1- (1/2)^n )