MHB Sequence of b_{k} with Explicit Formula: Proving by Math Induction

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The sequence defined by the recurrence relation b_k = b_{k-1}/2 + b_{k-1} with b_0 = 1 leads to confusion regarding its explicit formula. A proposed explicit formula is b_n = 2/(n+2), but this does not match the calculated terms of the sequence. Clarification is needed on whether the correct recurrence should be b_k = 3b_{k-1}/2 or b_k = b_{k-1}/(2 + b_{k-1}), as both yield different sequences. The accurate sequence appears to be {1, 1/3, 1/7, 1/15, ...}, suggesting a formula of 1/(2^{n+1} - 1). Further discussion is required to establish the correct explicit formula and its proof by mathematical induction.
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$$b_{k} = b_{k - 1}/2 +b_{k-1} $$
$$b_{0} = 1 $$

What would be the sequence for this expression, I calculated it to be 1, 2/3, 2/5, 2/7 ...

Is it right?

My explicit formula is $$b_{n} = 2/n+2 $$

What would be the explicit formula in your view and how can that formula be proved by mathematical induction? thanks
 
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First, what is the formula defining the sequence? You wrote $b_k= b_{k-1}/2+ b_{k-1}$ but since that would be more simply written as $3b_{k-1}/2$ I wonder if you did not intend $b_k= \frac{b_{k-1}}{2+ b_{k-1}}$.

Assuming what you wrote was correct, since $b_0= 1$, $b_1= 1/2+ 1= 3/2$, $b_2= (3/2)/2+ (3/2)= 3/4+ 3/2=3/4+ 6/4=10/4=5/2$... That is not what you write so apparently that is not what you intend.

With $b_k= \frac{b_{k-1}}{2+b_{k-1}}$ and $b_0=1$, $b_1= \frac{1}{2+ 1}= \frac{2}{3}$, $b_2= \frac{\frac{2}{3}}{2+ \frac{2}{3}}= \frac{\frac{2}{3}}{\frac{6}{3}+ \frac{2}{3}}= \frac{\frac{2}{3}}{\frac{8}{3}}= \frac{2}{3}\frac{3}{8}= \frac{2}{8}= \frac{1}{4}$.

That is also not what you have so, no, you are not correct.

I can't respond to the rest of you questions because I do not know whether you intend $b_k= \frac{b_{k-1}}{2}+ b_k= \left(\frac{3}{2}\right)b_{k-1}$ or $b_k= \frac{b_{k-1}}{2+ b_{k-1}}$.
 
$b_k = \dfrac{b_{k-1}}{2+b_{k-1}}$

$\displaystyle \bigg\{ 1, \dfrac{1}{3}, \dfrac{1}{7}, \dfrac{1}{15}, … , \dfrac{1}{2^{n+1}-1} , … \bigg \}$
 
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