Question

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

Answer 1

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 )