$n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$

Question

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?

Answer 1

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.