TITLE: The integer part of a sequence
QUESTION [7 upvotes]: I was trying to solve the following sum:
$$a_1=\sqrt[3]{24}$$
$$a_{n+1}=\sqrt[3]{(a_n+24)},n\ge1$$
$Find\,the\,integer\,part\,of\,a_{100}$
Source: ISI B Math 2012 paper

I proceeded in this manner:
$$a_1^3=24$$
$$a_2=\sqrt[3]{a_1+a_1^3}$$
$$a_2^3=a_1+a_1^3$$
similarly,
$$a_3^3=a_2+a_1^3=a_1+2a_1^3$$
$$a_4^3=a_1+3a_1^3$$
$$\ldots$$
$$a_{100}^3=a_1+99a_1^3$$
Which should give me the answer,
But I am encountering a dilemma in the part where I have to bring down the number to its third root before applying the greatest integer function.
Would appreciate a bit of help with this/ a new method of doing the same.

REPLY [11 votes]: $a_1<3$ then $a_2=\sqrt[3]{a_1+24}<\sqrt[3]{3+24}=3$.
Prove that $a_n<3$ by induction.
Suppose that for $n=k$: $a_k<3$ then $a_{k+1}=\sqrt[3]{a_k+24}<\sqrt[3]{3+24}=3$  . So $a_{100}<3$ but  $a_{100}>2$ that's why $[a_{100}]=2$.