Consider the given series.
21/4.41/8.81/16.161/32......
S=21/4.22/8.23/16.24/32......
S=2(1/4+2/8+3/16+4/32+.......)
S=21/4(1+2/2+3/4+4/8+.......)
Let
T=1+22+34+48+..... (Infinity AGP series with a=1, r=1/2, d=1)
Therefore,
T=a1−r+dr(1−r)2
T=11−12+1×12(1−12)2
T=112+12(12)2
T=2+2=4
Therefore,
S=21/4×4
S=21=2
Hence, this is the answer.