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.)

When you combined the fractions you missed the minus sign in front of the first fraction, the correct step starting from your second row is
$$3-\frac{2k+3}{2^k}+\frac{2(k+1)-1}{2^{k+1}}=3-\frac{(2k+3)(2^{k+1})-(2(k+1)-1)(2^{k})}{2^{k+1}2^{k}}.$$
To think a bit more easily why this is so just note that
$$-a+b=-a-(-b)=-(a-b).$$
PS. It is worth considering when dealing with fractions of similar denominators to combine them using stuff like
$$\frac{2k+3}{2^{k}}=\frac{2(2k+3)}{2^{k+1}}$$
