2+5+8+...+(3n-1)=(3n^2+n)/2

[asya]

how do i solve this?
2+5+8+...+(3n-1)=(3n^2+n) /2..................... edited .................. those parentheses ( ) are super-important

 

[Dr.Peterson]

how do i solve this?
2+5+8+...+(3n-1)=3n^2+n /2

Do you mean, how do you prove that 2+5+8+...+(3n-1)=(3n^2+n) /2 for all positive integers n?

That depends on what you have learned, and the goal of the proof. You might do it by induction, or by applying a formula you have learned (e.g. for arithmetic series), or various other ways. Which way would you like? Where are you stuck?

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[Deleted member 4993]

how do i solve this?
2+5+8+...+(3n-1)=(3n^2+n) /2

You were asked to "prove" an identity - not "solve it.

Do you know how to prove:

1 + 2 + 3 + ............................. + (n-2) + (n-1) + (n) = n * (n+1)/2

You have to use the same process.

Start with:

S_n = 2 + 5 + 8 + ..........+ (3n - 7) + (3n - 4) + (3n-1)

 

[pka]

\(\begin{gathered}
\sum\limits_{k = 1}^n {(3k - 1)} = 3\sum\limits_{k = 1}^n k - \sum\limits_{k = 1}^n1 \\
= 3\frac{{n(n + 1)}}{2} - n \\
= \frac{{3{n^2} + 3n}}{2} - n \\
= \frac{{3{n^2} + n}}{2} \\
\end{gathered} \)

 

[Deleted member 4993]

\(\begin{gathered}
\sum\limits_{k = 1}^n {(3k - 1)} = 3\sum\limits_{k = 1}^n k - \sum\limits_{k = 1}^n1 \\
= 3\frac{{n(n + 1)}}{2} - n \\
= \frac{{3{n^2} + 3n}}{2} - n \\
= \frac{{3{n^2} + n}}{2} \\
\end{gathered} \)

You had said:

Are we now giving out answers?

https://www.freemathhelp.com/forum/threads/probability-of-letters-in-password.124027/#post-503350 ...........[post#9]

 

[pka]

You had said:

https://www.freemathhelp.com/forum/threads/probability-of-letters-in-password.124027/#post-503350 ...........[post#9]

In this case, yes.

 

[Dr.Peterson]

@asya, I hope at some point you will tell us something about the context, as there are so many ways to approach this, we have no idea which answer might be suitable. The two specific answers (a hint and a full proof) are both different from the two approaches I explicitly mentioned, being part of my "other".