# How do you use summation notation to expression the sum 32+24+18+...+10.125?

Mar 9, 2017

${\sum}_{n = 0}^{4} 32 \cdot {\left(\frac{3}{4}\right)}^{n}$

#### Explanation:

$32 + 24 + 18 + \ldots + 10.125$

The initial pattern is:

$32 \cdot {\left(\frac{3}{4}\right)}^{0} + 24 \cdot {\left(\frac{3}{4}\right)}^{1} + 18 \cdot {\left(\frac{3}{4}\right)}^{2} + \ldots + 10.125$

And therefore:

$10.125 = 32 \cdot {\left(\frac{3}{4}\right)}^{n}$

$\implies n = \ln \frac{\frac{10.125}{32}}{\ln} \left(\frac{3}{4}\right) = n = 4$

So it's:

$32 \cdot {\left(\frac{3}{4}\right)}^{0} + 32 \cdot {\left(\frac{3}{4}\right)}^{1} + 32 \cdot {\left(\frac{3}{4}\right)}^{2} + \ldots + 32 \cdot {\left(\frac{3}{4}\right)}^{4}$

We can say then that the series sum is:

${\sum}_{n = 0}^{4} 32 \cdot {\left(\frac{3}{4}\right)}^{n}$