The easiest way to get the sum is to determine the first three terms using the given formula:
[tex]\text{ A}_n\text{ = }(\frac{8}{25})(\frac{5}{2})^{n\text{ - 1}}[/tex]We get,
First term, n = 1
[tex]\text{ A}_1\text{ = }(\frac{8}{25})(\frac{5}{2})^{n\text{ - 1}}\text{ = }(\frac{8}{25})(\frac{5}{2})^{1\text{ - 1}}\text{ = }(\frac{8}{25})(\frac{5}{2})^0[/tex][tex]\text{ A}_1\text{ = (}\frac{8}{25})(1)\text{ = }\frac{8}{25}[/tex]Second term, n = 2
[tex]\text{ A}_2\text{ = }(\frac{8}{25})(\frac{5}{2})^{2\text{ - 1}}\text{ = }(\frac{8}{25})(\frac{5}{2})^1[/tex][tex]\text{ A}_2\text{ = }(\frac{8}{25})(\frac{5}{2})^{}\text{ = }\frac{40}{50}\text{ = }\frac{20}{25}[/tex]Third term, n = 3
[tex]\text{ A}_3\text{ = }(\frac{8}{25})(\frac{5}{2})^{3\text{ - 1}}\text{ = }(\frac{8}{25})(\frac{5}{2})^2[/tex][tex]\text{ A}_3\text{ = }(\frac{8}{25})(\frac{25}{4})\text{ = }\frac{200}{100}\text{ = }\frac{50}{25}[/tex]Since the first three terms are already like terms (the same denominator), let's proceed on adding them.
[tex]\text{ Sum = }A_1+A_2+A_3_{}[/tex][tex]\text{ Sum = }\frac{8}{25}\text{ + }\frac{20}{25}\text{ + }\frac{50}{25}\text{ = }\frac{8\text{ + 20 + 50}}{25}\text{ = }\frac{78}{25}[/tex]Therefore, the sum of the first three terms is 78/25.
