Can someone please write out steps as to how they'd solve a problem like this? (Algebra) Thanks!

Question

anonymous · Answer

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zepdrix · Answer

$$\large \color{royalblue}{\frac{2n}{3}}$$

Under the `sigma` it says $\large n=1$
This is telling us that for our first term, we plug in $\large \color{orangered}{1}$ for our n.

$$\large \color{royalblue}{\frac{2\cdot\color{orangered}{1}}{3}} \qquad = \qquad \frac{2}{3}$$

Our next term will be $\large n=\color{orangered}{2}$.
$$\large \color{royalblue}{\frac{2\cdot\color{orangered}{2}}{3}} \qquad = \qquad \frac{4}{3}$$

The `sigma` is telling us to ADD together each of these successive terms.
$$\large \frac{2}{3}+\frac{4}{3}+....$$

But when do we stop adding?
The number ABOVE the sigma tells us when to stop.


So based on that information, let's see if you can figure out part a.
$\large n=1$ represents the first term.
$\large n=2$ the second.
and $\large n=8$ the last term.

Hmm so how many terms will there be?

zepdrix · Answer

Keep in mind, we're counting by whole numbers*