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

[hkhgg]

$$b_{k} = b_{k - 1}/2 +b_{k-1} $$
$$b_{0} = 1 $$

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

 

[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]

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

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