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I'm supposed to rewrite the sum 1,3,7,15,31...1023 using summation notation, then find the sum. I can tell the value being added begins with 2 then doubles to 4,8,etc. But I'm not sure how to write it in summation notation, I figure it's some sort of polynomial, but I don't know how I would be able to solve for that or anything. Any help would be appreciated.

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    $\begingroup$ If the value being added is supposed to double, then the fifth term should be 31, not 25. $\endgroup$
    – user694818
    Commented Aug 13, 2019 at 22:00
  • $\begingroup$ Sorry, fixed it! $\endgroup$
    – JCHLW
    Commented Aug 13, 2019 at 22:02

1 Answer 1

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As you can see, and you suggest in your question text, each term is the next power of $2$ less $1$. Thus, since $1023 = 2^{10} - 1$, the sum you want is

$$\begin{equation}\begin{aligned} \sum_{i=1}^{10} (2^i - 1) & = \sum_{i=1}^{10} 2^i - \sum_{i=1}^{10}1 \\ & = (2^{11} - 2) - (10) \\ & = 2046 - 10 \\ & = 2036 \end{aligned}\end{equation}\tag{1}\label{eq1}$$

Note I got $\sum_{i=1}^{10} 2^i = 2^{11} - 2$ using the Geometric series sum of $a\left(\frac{r^n - 1}{r - 1}\right)$ where $a = 2$ is the first term, $n = 10$ is the number of terms, and $r = 2$ is the common ratio. Thus, you get $2(2^{10} - 1) = 2^{11} - 2$.

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  • $\begingroup$ $2^{11}-2=2048-2=204\color{red}6$ $\endgroup$ Commented Aug 13, 2019 at 22:11
  • $\begingroup$ @MatthewDaly Thanks for pointing it out. I've already fixed the problem. $\endgroup$ Commented Aug 13, 2019 at 22:12
  • $\begingroup$ eep, thank you for correctly identifying what the problem was.... $\endgroup$
    – user694818
    Commented Aug 13, 2019 at 22:12
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    $\begingroup$ @J.W.Tanner Thanks for point out that error. I've fixed it now. $\endgroup$ Commented Aug 13, 2019 at 22:12

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