Consider the sequence $a_1 = 24^{1/3}$ $a_{n+1} = (a_n + 24)^{1/3},n ≥ 1.$ Then what is the integer part of $a_{100}$?

Question

QUESTION: Consider the sequence

$a_1 = 24^{1/3}$ $a_{n+1} = (a_n + 24)^{1/3},n ≥ 1.$

Then what is the integer part of $a_{100}$ ?

MY APPROACH: I tried this one really hard but couldn't get the trick.. I used log, but that doesn't really help and the problem becomes more and more complex, so I am avoiding a confusing solution here..

Then I tried by defining a function say $$f(x)=(x+24)^\frac{1}3$$ Therefore by computing the derivative of $f$ we find that the rate at which the function increases, decreases with increase in x. Which also is quite clear from intuition. But I could not apply the result to solve the problem.

Can we form a recursive series for it? Any help will be much appreciated. Thank you so much.

Answer 1

Hint: Prove by induction that $2 < a_n < 3$ for all $n$.

Answer 2

If $x<3$, then $f(x)<(3+24)^{1/3}=3$. If $x\geq 0$, then $f(x)\geq(0+24)^{1/3}>2$.

So what can you say about $a_{100}=f^{100}(0)$?

Answer 3

given , $a_1= 24^{1/3}$

$\implies $$a_1 <3$ as $27^{1/3}$ is 3

$a_{n+1}=(a_n+24)^{1/3} $

$a_2 = (a_1+24)^{1/3}$

since $a_1<3 $

$\implies $$(a_1+24)^{1/3}$ <3

$\implies $$a_2<3$

similary $a_3<3$, $a_4<3$ and so on

$\Rightarrow$ $a_n<3$ $\Rightarrow$ $a_{100}$<3 ----1

Now, for any value of $a_n$(+ve,<3) in $a_{n+1}=(a_n+24)^{1/3}$ the expression is always greater than 2

$\Rightarrow$ 2<$a_n$<3

$\Rightarrow$ 2<$a_{100}$<3

Therefore , the integral part of $a_{100}$ is 2.