an = 2^(n-1)
b1=a1=1
bn = b1+(n-1)d
= 1+(n-1)d
bm= am
1+(m-1)d = 2^(m-1)
d = [2^(m-1) -1]/( m-1)
Sm=a1+a2+...+am= (2^m) -1
Tm =(b1-1/2) + (b2-1/2)+...+(bm-1/2)
= (1/2) + (1/2+d)+...+(1/2+(m-1)d)
= m/2 + m(m-1)d/2
= m/2 + (m/2)[2^(m-1) -1]
= m.2^(m-2)
m =1
S1=1,T1=1/2
S1 > T1
m=2
S2 = 3 ,T2=2
S2 > T2
m=3
S3 = 7,T3=6
S3 > T3
m ≥4
Tm = m.2^(m-2)
≥ 4 .2^(m-2)
= 2^m
> 2^m - 1
= Sm
ie
Sn < Tn :n≤3
Tn > Sn ; n≥4