$n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ Prove that $n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ is given by 
$$T_n =\left[\sqrt{2n+\frac{1}{2}}\right]$$ where $[.]$ is Floor function
My Try:
Its clear that $1$st Term is $1$, $3$rd Term is $2$, $6$th term is $3$ and so on
We have Triangular numbers as $1,3,6,10,...$ whose $m$th term is given by $\frac{m(m+1)}{2}$
Hence for the given sequence
$$T_{\frac{m(m+1)}{2}}=m$$
Letting $$\frac{m(m+1)}{2}=n$$ we get Quadratic in $m$ as
$$m^2+m-2n=0$$ So $m$ takes $\frac{-1\pm \sqrt{1+8n}}{2}$ and since $m$ is a positive integer we have
$$m=\frac{-1+\sqrt{1+8n}}{2}$$
hence
$$T_n=\frac{-1+\sqrt{1+8n}}{2}$$
how to proceed further? 

Note that $$T_n=\frac{-1+\sqrt{1+8n}}{2}$$ is true only if $$n=m(m+1)/2.$$ Now you have to define $T_n$ for other values of $n$. Since terms repeat between two such triangular numbers and you know the value of  $T_n$ for $n=m(m+1)/2$ the rest of the solution should be obvious.    
