Question

.

Answer 1

13.5

Answer 2

\[\sum_{k=-2}^{27}3^k\]
\[=9+2+\sum_{k=0}^{27}(\frac{1}{3})^k\]

Answer 3

That should be -k in the first equation.

Answer 4

well it looks like a geometric series where the 1st term a = 9 and the common ratio r = 1/3, and the number of terms n = 30


so using the sum of a geometric series formula

\[S_{n} = \frac{a(1 - r^n)}{1 - r}\]

just substitute and then evaluate

Answer 5

\[r^0+r^1+r^2+r^3....+r^n=\frac{1-r^{n+1}}{1-r}\]