Big bases

Base 2: the length of the number is about $log_2(10^{100})$ , and each digit averages at (0+1)/2. So the digit sum should be near $log_2(10^100)/2=100log_2(10)/2.$

Base 3: The length of the number is $~log_3(10^100)$ , and each digit averages at $(0+1+2)/3= 1$ . So the digit sum should be near $100log_3(10).$

Base N: The length of the number is $~log_N(10^100)$ , and each digit averages at $N(N-1)/(2N)= (N-1)/2$ . So the digit sum should be near $100(N-1)log_N(10)/2=(50ln(10))(N-1)/ln(N)$ .

Since N-1 grows faster than $ln(N)$ , the digit sum is generally increasing.

Graph of digit sum of 10^100: http://www.wolframalpha.com/input/?i=digit+sum+of+googol+in+base+2&lk=2

Let's say the 100 digit number is 10^100, which in binary form has about 300 digits. Hence, the maximum sum of digits is a about 300, or a number about 3 digits long. However, that same 100 digit number in some 100 digit base, then any one digit of it is a number that could be about 100 digits long. The same observation can be made with numbers less than 100 digits.