Question 972087


Hello,


Using the rule {{{(p/q)^n =  p^n/q^n}}} we have (2/3)^n 

where n represents the exponents 1, 2, 3, 4, +... +, 100. So for,


 *[illustration approximation.jpg]


The sum of (2/3) + (2/3)^2 + (2/3)^3 + (2/3)^4 +....+ (2/3)^100 will be as follows:


*[illustration sum.jpg]


and the decimal approximation is 1.99999...


Your answer 2-2(2/3)^100 is correct.