Do you mean, how do you prove that 2+5+8+...+(3n-1)=(3n^2+n) /2 for all positive integers n?how do i solve this?
2+5+8+...+(3n-1)=3n^2+n /2
You were asked to "prove" an identity - not "solve it.how do i solve this?
2+5+8+...+(3n-1)=(3n^2+n) /2
You had said:\(\begin{gathered}
\sum\limits_{k = 1}^n {(3k - 1)} = 3\sum\limits_{k = 1}^n k - \sum\limits_{k = 1}^n1 \\
= 3\frac{{n(n + 1)}}{2} - n \\
= \frac{{3{n^2} + 3n}}{2} - n \\
= \frac{{3{n^2} + n}}{2} \\
\end{gathered} \)
Are we now giving out answers?
In this case, yes.You had said:
https://www.freemathhelp.com/forum/threads/probability-of-letters-in-password.124027/#post-503350 ...........[post#9]