I was trying to solve the following sum:

$$a_1=\sqrt[3]{24}$$ $$a_{n+1}=\sqrt[3]{(a_n+24)},n\ge1$$ $Find\,the\,integer\,part\,of\,a_{100}$

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.

  • $\begingroup$ Haven't you missed something in question? $\endgroup$ – Mostafa Ayaz Jan 6 '18 at 18:29
  • $\begingroup$ No I haven't missed anything $\endgroup$ – Saurav Goyal Jan 6 '18 at 18:40
  • $\begingroup$ You've used the wrong value for $a_2$ when you calculated $a_3$ - you put in the value for $a_2^3$ instead... (just in case you wanted to know where you went wrong on your approach) $\endgroup$ – Chris Jan 6 '18 at 23:23
  • $\begingroup$ Yes right. This was helpful $\endgroup$ – Saurav Goyal Jan 7 '18 at 19:12

$a_1<3$ 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 $[a_{100}]=2$.


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