Series/Sequences help *question attached below* will give medal

Question

itiaax · Answer

Can I have a walkthrough with this?

ganeshie8 · Answer

you're given $a_1 = 1$
so the statement  $\large 1\le a_n\le 8  $ is true for $n=1$

ganeshie8 · Answer

next assume the given statement is true for some $n=k$ and prove it will be true for  $n = k+1$ too

ganeshie8 · Answer

assuming  $\large 1\le a_k \le 8$, you need to prove :
$$\large 1\le a_{k+1}\le 8$$

ganeshie8 · Answer

$$\large 1\le a_{k}\le 8$$

take cuberoot through out you get :

$$\large 1\le \sqrt[3]{a_{k}}\le 2$$

multiply 2 through out :

$$\large 2\le 2\sqrt[3]{a_{k}}\le 4$$

add 4 through out :

$$\large 6\le4+ 2\sqrt[3]{a_{k}}\le 8$$

ganeshie8 · Answer

notice that expression in the middle is same as  $\large  a_{k+1}$

ganeshie8 · Answer

so we are done :$$\large 6\le a_{k+1}\le 8$$

itiaax · Answer

Allow me to process and understand everything

ganeshie8 · Answer

take ur time

itiaax · Answer

Got it! Thanks! @ganeshie8 :)

ganeshie8 · Answer

you're welcome :)