ith term of$~\left\{a\right\}=\left\{1,1,3,1,3,5,1,3,5,7,1,3,5,7,9,1,\cdot\cdot\cdot\right\}~$such that$~a_{i}=17~$for j times appearnce

Solution 1:

As we suppose$~i\in\mathbb{N}~$,

$~2i-1~$of coursely takes an odd number and this means that,

$$\text{ith term}=2i-1\tag{1}$$

$$\therefore~~i=9~~\Leftrightarrow~~\text{ith term}=17\tag{2}$$

Moreover, I wrote the following.

Of course as we enumurate$~1,3,5,7,\cdot\cdot\cdot~$,9th term of this sequence takes 17.

We nextly focus on mth appearance of 17 of$~\left\{a\right\}~$

$$17_{m}:=\text{17 with mth appearnce}~~~\left(m\in\mathbb{N}\right)\tag{3}$$

$~m+8~$th groups'$~9~$th term$~~~\Leftrightarrow~~17_{m}~$

$$\sum_{i=1}^{m+7}i=\frac{1}{2}\cdot\left(m+7\right)\left(1+\left(m+7\right)\right)\tag{4}$$

$$=\underbrace{\frac{\left(m+7\right)\left(m+8\right)}{2}}_{\text{number of terms of}~\left\{a\right\}~\text{till end of}~m+7~\text{th group}}\tag{5}$$

$$\therefore~~n=\frac{\left(m+7\right)\left(m+8\right)}{2}+9\tag{6}$$

Finished.