Solution: (1) According to the meaning of the question,12 =1× 23+1× 221× 20, then I (12) = 2;

(2) 127= 1×26+ 1×25+ 1×24+ 1×23+ 1×22+ 1×2 1+ 1×20,

Let 64≤n≤ 126, where n is an integer;

Then n =1× 26+a/kloc-0 /× 25+a2× 24+a3× 23+a4× 22+a5× 21+a6× 20,

When six numbers in a 1, a2, a3, a4, a5 and a6 are all 0 or 1 and none of them are 1, there are C60 cases, that is, C60i (n) = 5;

When one of them is 1, there are C60 cases, that is, C60i (n) = 5;

When two of them are 1, there are C62 cases, that is, C62i (n) = 4;

…

∑n = 64 1272 I(n)= c 6026+c 6 1×25+c62×24+c63×23+c64×22+c65×2+ 1 =(2+ 1)n = 36，

Similarly: ∑n=32632I(n)=35,

…

∑n=232I(n)=3 1，

2I( 1)= 1；

Then ∑ n =11272i (n) =1+3+32+…+36 = 37-13-1=1093;