Question

the summation of 48 times one fourth to the i minus 1 power, from i equals 1 to infinity.

Answer 1

This is a classic sum of geometric series : \[\sum_{i=1}^{\infty} 48 (1/4)^{i-1} = \lim_{n \rightarrow \infty} \sum_{j=0}^{n} 48 (1/4)^{j} = \lim_{n \rightarrow \infty} 48 \frac{ 1-(1/4)^{n+1} }{1- (1/4) }\] I'll let you finish.