Question

Find the least value of n for which the sum 1+3+3^2..To n terms is greater than 7000

Answer 1

$S_n=1+3+3^2+\ldots+3^{n-1}>7000\\a=1\\r=3\\\\S_n=\dfrac{a(1-r^n)}{1-r}\\S_n=\dfrac{1-3^n}{1-3}=-\dfrac{1-3^n}{2}\\\\-\dfrac{1-3^n}{2}>7000\\1-3^n<-14000\\3^n>14001\\n>\log_314001(\approx8.7)$

Since $n$ must be natural, the least value is 9.