`sum_(n = 1)^7 4^(n-1)` Find the sum of the finite geometric series.

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` .