Answer: The required sum is 259.
Step-by-step explanation: We are given to find the sum of the following geometric series :
[tex]\sum_{i=1}^{4}6^{i-1}.[/tex]
The given geometric series can be written as :
[tex]0+6+6^2+6^3.[/tex]
We know that
the sum of a geometric series up to n terms with first term a and common ration r is given by
[tex]S=\dfrac{a(r^n-1)}{r-1}.[/tex]
in the given series, we have
[tex]a=1,~~r=\dfrac{6}{1}=\dfrac{6^2}{6}=\dfrac{6^3}{6^2}=6.[/tex]
Therefore, the sum up to 4 terms will be
[tex]S_4=\dfrac{a(r^4-1)}{r-1}=\dfrac{1(6^4-1)}{6-1}=\dfrac{1296-1}{5}=\dfrac{1295}{5}=259.[/tex]
Thus, the required sum is 259.
