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$$\sum_{i=1}^n \frac{2i-1}{2^i} = 3 - \frac{2i+3}{2^i}$$

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.)

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    $\begingroup$ Hint: How would you add $\frac{1}{8} + \frac{3}{16}$? $\endgroup$ Commented Feb 16, 2023 at 0:25
  • $\begingroup$ $$ 3 - \frac{(2k+3)(2^{k+1}) - (2(k+1)-1)(2^k)}{(2^k)(2^{k+1})}$$ $\endgroup$ Commented Feb 16, 2023 at 0:27
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    $\begingroup$ Notice that : $$\dfrac{a}{2^k} + \dfrac{b}{2^{k + 1}} = \dfrac{2 a + b}{2^{k + 1}}$$ $\endgroup$ Commented Feb 16, 2023 at 0:28

3 Answers 3

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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 $$\frac{2k+3}{2^{k}}=\frac{2(2k+3)}{2^{k+1}}$$

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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.

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  • $\begingroup$ How do I get rid of the extra 4k? $\endgroup$ Commented Feb 16, 2023 at 1:13
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    $\begingroup$ Welcome to Math.SE. It's nice that you spotted OP's mistake in the original statement, but without correcting their mistake when they added their fractions, I don't know that they would have been able to get the answer from just your help. $\endgroup$ Commented Feb 16, 2023 at 14:26
  • $\begingroup$ @James Remember, the first fraction is negative, you are subtracting it from the total, not adding it. With that in mind, add the numerators together and see what happens. $\endgroup$ Commented Feb 16, 2023 at 18:10
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Your request that we not reveal the answer is a healthy attitude, as it shows that you are trying to understand the math, rather than just getting the answer. Taking this further, rather than telling you what your error is, I'll share a tactic that may help you find such errors in general. This tactic is substituting in a particular value. For instance, substituting in $k=1$, we have

$$\sum_{i=1}^1 \frac{2i-1}{2^i}+\frac{3}{4}$$

$$3 - \frac{5}{2} + \frac{3}{4}$$

$$ 3 - \frac{5 \cdot 4 +3\cdot 2}{2 \cdot 4}$$

$$ 3 - \frac{5 \cdot 4 +3\cdot 2}{2 \cdot 2 \cdot 2}$$

If you're still having trouble seeing the error, or at least on what step you made the error, converting to decimals may help.

$$\sum_{i=1}^1 \frac{2i-1}{2^i}+0.75$$

$$3 - 2.5 + 0.75$$

$$ 3 - 3.25$$

$$ 3 - 3.25$$

Also, two notes:

$i$ is a bound variable, so it shouldn't appear on the LHS. Instead you should have $n$.

When adding or subtracting fractions, the math is generally simpler if you use the lowest common denominator, rather than multiplying the denominators together. $2^k$ is a factor of $2^{k+1}$, so the lowest common denominator is just $2^{k+1}$.

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