# Question

• MHB
hkhgg
$$\displaystyle b_{k} = b_{k - 1}/2 +b_{k-1}$$
$$\displaystyle b_{0} = 1$$

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 $$\displaystyle b_{n} = 2/n+2$$

What would be the explicit formula in your view and how can that formula be proved by mathematical induction? thanks

HOI
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 $b_k= \frac{b_{k-1}}{2+ b_{k-1}}$.

skeeter
$b_k = \dfrac{b_{k-1}}{2+b_{k-1}}$

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