The AM of the series 1,2,4,8,16,…,2n is
2n−1n
2n+1−1n+1
2n+1n
2n−1n+1
The required AM is X¯=1+2+22+23+⋯+2nn+1 =2n+1−1(n+1)(2−1)=2n+1−1n+1