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How can this sum be rewritten like this

$1 + 1 + 2 + 2 + · · · + l + l + (l + 1) + (l + 1) + l + ε$

$= 2\sum\limits_{i=1}^{l+1}i+l+ε$

$= (l+1)(l+2)+l+ε $

First of all I dont even understand how we can go from the first step to the second. To make this a bit more clear for myself I tried plugging in numbers into the second step like this. But I'm pretty its wrong-

$2\sum\limits_{i=1}^{l+1}i+l+ε$

$=2\sum\limits_{i=1}^{l+1}1+l+ε + 2 + l + ε $

...

JangoCG
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2 Answers2

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You seem to be including the $l$ and $\varepsilon$ inside the sum as if it were $$\sum \left( i + l + \varepsilon \right)$$

But this is not the situation.

The first line has two copies of every number from $1$ to $l+1$, and a lone copy of $l + \varepsilon$. To put this into a summation form, we can write $$\left(\sum_{i=1}^{l+1} 2i \right) + l+ \epsilon$$

Using summation properties, we can factor out the $2$ from the sum: $$2 \left( \sum_{i=1}^{l+1} i \right) + l+ \epsilon$$

This is how the second line should read.

Can you get to the third line by evaluating the sum?

WaveX
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  • Thanks a lot I finally understood and could compute the third step all by myself and get the right solution! However how did you know that only $i$ is included in the sum? I copied my formula straight from my professors slides. Do there have to be parentheses if more then one variable is inside a sum? – JangoCG Nov 20 '20 at 16:01
  • If it were in the parenthesis, we would have many more copies of $l+\varepsilon$ in the first row. But we only have the one, shown at the end of the first line – WaveX Nov 20 '20 at 16:22
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Using the sum of first $n$ numbers formula $(1+2+...+ n = \frac {n(n+1)}{2})$,

$1+2+...+(l+1)= \sum\limits_{i=1}^{l+1}i = $ ${\frac {(l+1)(l+2)}{2}} \implies 2\sum\limits_{i=1}^{l+1}i=(l+1)(l+2)$ so that

$2(1+2+...+(l+1))=1+1+2+2+...+(l+1)+(l+1)=(l+1)(l+2)$ so that

$1+1+2+2+...+(l+1)+(l+1)+l+\epsilon=(l+1)(l+2)+l+\epsilon$

Derek Luna
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