Aditya Gupta
Last Activity: 6 Years ago
since (1+x)(1+x2)(1+x4).......(1+x128)=∑xr where r=0 and limit is n, all we gotta do is to find the degree of the polynomial on LHS, which is equal to 1+2+4+....128
this is a GP with common ratio 2, so sum is:
S=1*(2^8 – 1)/(2 – 1)
so n=S=2^8 – 1
n= 255