Question

Please help me understand this, 2 algebra questions

Answer 1

\[\sum_{n=1}^{5}(3n+1)\]

Answer 2

\[\sum_{n=1}^{8}2n/3\]

Answer 3

I have to figure out the Number of terms, the first term, the last term. Lastly having to evaluate the sum

Answer 4

\(\large \sum_{{\color{red}{ n}}=1}^{5}(3{\color{red}{ n}}+1)\implies
\begin{array}{llll}\
[3({\color{red}{ 1}})+1]+\\\
[3({\color{red}{ 2}})+1]+\\\
[3({\color{red}{ 3}})+1]+\\\
[3({\color{red}{ 4}})+1]+\\\
[3({\color{red}{ 5}})+1]+\\
\hline\\
\end{array}\)

Answer 5

that's pretty much what the \(\Sigma\) notation means

Answer 6

Ohhhh, That makes alot more sense, so how do I determine the number of terms?

Answer 7

well, \(\Large \Sigma_{n=\textit{start "n" at this value}}\qquad \Sigma^{\textit{end "n" at this value}}\)

Answer 8

and \(\Sigma\) means SUMMATION, or grab all those terms and SUM THEM up

Answer 9

So that means, n=1 is the amount of terms?

Answer 10

well, the amount of terms will be
the amount of values between START and END points

in this case n = 1 is the start, so 1
the end is n = 5, so 5 is the end

between 1..5 you have that many terms

Answer 11

so in this case, it'd be 5 terms, since changing "n" from 1 till 5, will yield 5 terms :)