Precalculus, Chapter 9, 9.3, Section 9.3, Problem 55

The given geometric series is:
sum_(n=1)^7 4^(n-1)
Take note that if the geometric series has a form:
sum_(n=1)^n a_1 * r^(n-1)
its finite sum is:
S_n=a_1*(1-r^n)/(1-r)
Rewriting the given sigma notation in exact form as above, it becomes:
sum_(n=1)^7 4^(n-1)=sum_(n=1) 1 * 4^(n-1)
From here, it can be seen that the values of the first term and the common ratio are a1=1 and r=4.
Plugging in the values of a1 and r to the formula of Sn, the sum of the first seven terms of the geometric sequence is:
S_7=1*(1-4^7)/(1-4) = 5461

Therefore, sum_(n=1)^7 4^(n-1) = 5461 .