What am i doing wrong for this induction problem? $$\sum_{i=1}^n \frac{2i-1}{2^i} =  3 - \frac{2i+3}{2^i}$$
This equation has been giving me a lot of trouble, and I don't know what I am doing wrong for this question. I'm trying to get this: $$  3 - \frac{2(k+1)+3}{2^{k+1}}$$
Here is my work:
$$\sum_{i=1}^k \frac{2i-1}{2^i}+\frac{2(k+1)-1}{2^{k+1}}$$
$$3 - \frac{2k+3}{2^k} + \frac{2(k+1)-1}{2^{k+1}}$$
$$ 3 - \frac{(2k+3)(2^{k+1})+(2(k+1)-1)(2^k)}{(2^k)(2^{k+1})}$$
$$ 3 - \frac{(2k+3)(2^{k})(2)+(2(k+1)-1)(2^k)}{(2^k)(2^{k})(2)}$$
Of course, I've tried other methods, but this is the farthest I've gotten. The base case checks out, and I know that this works out, but I just don't know where I've gone wrong. Can anyone tell me what I'm not doing? (This is for all values greater than or equal to 1 and Try not to reveal the answer, please.)

To be clear, I presume the statement you are trying to prove is $$ \sum_{i=1}^{n}\frac{2i - 1}{2^i} = 3 - \frac{2n + 3}{2^n}$$ as opposed to there being $i$'s in the fraction to the right. In this case, your first two steps were in the right direction, but remember that the form you'd like is $$3 - \frac{2(k + 1) + 3}{2^{k+1}}$$
So then with the form you have in your second line of work, $$3 - \frac{2k + 3}{2^k} + \frac{2(k+1) - 1}{2^{k+1}} $$ you can manipulate the fraction on the left in order to make it have the same denominator as the one on the right: $$3 - \frac{2k + 3}{2^k} + \frac{2(k+1) - 1}{2^{k+1}} $$ $$= 3 - \frac{2k + 3}{2^k} \left(\frac{2}{2}\right) + \frac{2(k+1) - 1}{2^{k+1}} $$
$$= 3 - \frac{4k + 6}{2^{k+1}} + \frac{2(k+1) - 1}{2^{k+1}} $$ And then you can get it to the desired form from there.
