# How do you solve this sigma problem?

## $\sum$3(5)n-1 There is an 8 above the sigma and n=1

May 11, 2018

$\textcolor{b l u e}{292968}$

#### Explanation:

Not sure of your notation here. I am assuming it is:

${\sum}_{n = 1}^{8} 3 \cdot {\left(5\right)}^{n - 1}$

We can recognise $3 \cdot {\left(5\right)}^{n - 1}$ as being the nth term of a geometric series:

$a {r}^{n - 1}$

Where:

$\boldsymbol{a}$ is the first term, $\boldsymbol{r}$ is the common ratio and $\boldsymbol{n}$ is the nth term:

The sum of a geometric series is given as:

$a \left(\frac{1 - {r}^{n}}{1 - r}\right)$

So plugging in the known values:

$3 \left(\frac{1 - {5}^{8}}{1 - 5}\right) = 3 \left(\frac{- 390624}{-} 4\right) = 3 \left(97656\right) = 292968$

${\sum}_{n = 1}^{8} 3 \cdot {\left(5\right)}^{n - 1} = 292968$

May 11, 2018

$532$

#### Explanation:

$\text{assuming } {\sum}_{n = 1}^{8} 3 \left(5\right) n - 1$

$\text{using the "color(blue)"summation blocks}$

•color(white)(x)sum_(r=1)^n 1=n

•color(white)(x)sum_(r=1)^n r=1/2n(n+1)

$\text{the question simplifies to}$

${\sum}_{n = 1}^{8} 15 n - 1$

$= 15 {\sum}_{n = 1}^{8} n - {\sum}_{n = 1}^{8} 1$

$= \frac{15}{2} \times \left(8 \times 9\right) - 8$

$= 540 - 8 = 532$