How do you find #S_n# for the geometric series #a_1=625#, r=3/5, n=5?
1 Answer
Mar 10, 2018
Explanation:
#"For a geometric series the sum to n terms is"#
#•color(white)(x)S_n=(a_1(1-r^n))/(1-r)color(white)(x)r!=1#
#"here "a_1=625" and "r=3/5#
#rArrS_n=(625(1-(3/5)^n))/(1-3/5)#
#color(white)(S_n)=3125/2(1-(3/5)^n)#
#"for "n=5#
#S_5=3125/2(1-(3/5)^5)#
#color(white)(S_5)=3125/2(1-243/3125)#
#color(white)(S_5)=3125/2xx2882/3125=1441#
