Post History
#4: Post edited
by
Moshi
·
2021-07-11T19:45:24Z (almost 3 years ago)
Fixed LaTeX
How does the change of variable $\color{red}{r↦n−r}$ transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum\limits_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$?
- I'm unskilled at performing algebra with Capita-sigma notation. [This comment by a deleted user](https://math.stackexchange.com/questions/2219125/prove-using-combinatorics-sum-limits-r-0n2n-r-binomnrn-22n?rq=1#comment4564672_2219125) alleges that the "change of variables $\color{red}{r↦n−r}$" will transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum\limits_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$.
I got stuck. If I replace $r$ with $\color{red}{n−r}$, then $\sum\limits_{r=0}^n 2^{n-r} \binom{n+r}{n}=2^{2n} \iff \sum\limits_{\color{red}{n−r}=0}^n 2^{n - \color{red}{(n−r)}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n} \iff sum \limits_{\color{red}{r=n}}^n 2^{\color{red}{r}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n}$. Can someone please rectify my MathJax?- Why can I bring $2^{2n}$ on the RHS into the indexed variable (representing each term of the sum)? On the LHS, $n$ are the lower and upper bounds of summation. So $n$ isn't a Free Variable. I don't know if I'm using "Free Variable" correctly, but what I mean is that because if the RHS were $2^{2x}$, then I can bring $2^{2x}$ into the indexed variable because $x$ never appears in the capital-sigma notation.
- I'm unskilled at performing algebra with Capita-sigma notation. [This comment by a deleted user](https://math.stackexchange.com/questions/2219125/prove-using-combinatorics-sum-limits-r-0n2n-r-binomnrn-22n?rq=1#comment4564672_2219125) alleges that the "change of variables $\color{red}{r↦n−r}$" will transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum\limits_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$.
- I got stuck. If I replace $r$ with $\color{red}{n−r}$, then $\sum\limits_{r=0}^n 2^{n-r} \binom{n+r}{n}=2^{2n} \iff \sum\limits_{\color{red}{n−r}=0}^n 2^{n - \color{red}{(n−r)}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n} \iff \sum\limits_{\color{red}{r=n}}^n 2^{\color{red}{r}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n}$
- Why can I bring $2^{2n}$ on the RHS into the indexed variable (representing each term of the sum)? On the LHS, $n$ are the lower and upper bounds of summation. So $n$ isn't a Free Variable. I don't know if I'm using "Free Variable" correctly, but what I mean is that because if the RHS were $2^{2x}$, then I can bring $2^{2x}$ into the indexed variable because $x$ never appears in the capital-sigma notation.
#3: Post edited
by
DNB
·
2021-07-11T06:55:44Z (almost 3 years ago)
How does the change of variable $\color{red}{r↦n−r}$ transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$?
- How does the change of variable $\color{red}{r↦n−r}$ transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum\limits_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$?
I'm unskilled at performing algebra with Capita-sigma notation. [This comment by a deleted user](https://math.stackexchange.com/questions/2219125/prove-using-combinatorics-sum-limits-r-0n2n-r-binomnrn-22n?rq=1#comment4564672_2219125) alleges that the "change of variables $\color{red}{r↦n−r}$" will transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$.- I got stuck. If I replace $r$ with $\color{red}{n−r}$, then $\sum\limits_{r=0}^n 2^{n-r} \binom{n+r}{n}=2^{2n} \iff \sum\limits_{\color{red}{n−r}=0}^n 2^{n - \color{red}{(n−r)}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n} \iff sum \limits_{\color{red}{r=n}}^n 2^{\color{red}{r}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n}$. Can someone please rectify my MathJax?
- Why can I bring $2^{2n}$ on the RHS into the indexed variable (representing each term of the sum)? On the LHS, $n$ are the lower and upper bounds of summation. So $n$ isn't a Free Variable. I don't know if I'm using "Free Variable" correctly, but what I mean is that because if the RHS were $2^{2x}$, then I can bring $2^{2x}$ into the indexed variable because $x$ never appears in the capital-sigma notation.
- I'm unskilled at performing algebra with Capita-sigma notation. [This comment by a deleted user](https://math.stackexchange.com/questions/2219125/prove-using-combinatorics-sum-limits-r-0n2n-r-binomnrn-22n?rq=1#comment4564672_2219125) alleges that the "change of variables $\color{red}{r↦n−r}$" will transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum\limits_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$.
- I got stuck. If I replace $r$ with $\color{red}{n−r}$, then $\sum\limits_{r=0}^n 2^{n-r} \binom{n+r}{n}=2^{2n} \iff \sum\limits_{\color{red}{n−r}=0}^n 2^{n - \color{red}{(n−r)}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n} \iff sum \limits_{\color{red}{r=n}}^n 2^{\color{red}{r}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n}$. Can someone please rectify my MathJax?
- Why can I bring $2^{2n}$ on the RHS into the indexed variable (representing each term of the sum)? On the LHS, $n$ are the lower and upper bounds of summation. So $n$ isn't a Free Variable. I don't know if I'm using "Free Variable" correctly, but what I mean is that because if the RHS were $2^{2x}$, then I can bring $2^{2x}$ into the indexed variable because $x$ never appears in the capital-sigma notation.
#2: Post edited
by
DNB
·
2021-07-11T06:25:35Z (almost 3 years ago)
How does the change of variable $\color{red}{r↦n−r}$" transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$?
- How does the change of variable $\color{red}{r↦n−r}$ transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$?
#1: Initial revision
by
DNB
·
2021-07-11T06:25:22Z (almost 3 years ago)
How does the change of variable $\color{red}{r↦n−r}$" transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$?
I'm unskilled at performing algebra with Capita-sigma notation. [This comment by a deleted user](https://math.stackexchange.com/questions/2219125/prove-using-combinatorics-sum-limits-r-0n2n-r-binomnrn-22n?rq=1#comment4564672_2219125) alleges that the "change of variables $\color{red}{r↦n−r}$" will transmogrify $\sum\limits_{r=0}^n2^{n-r}\binom{n+r}{n}=2^{2n}$ into $\sum_{k=0}^{n} \frac{\binom{2n-k}{n}}{2^{2n-k}}=1$.
I got stuck. If I replace $r$ with $\color{red}{n−r}$, then $\sum\limits_{r=0}^n 2^{n-r} \binom{n+r}{n}=2^{2n} \iff \sum\limits_{\color{red}{n−r}=0}^n 2^{n - \color{red}{(n−r)}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n} \iff sum \limits_{\color{red}{r=n}}^n 2^{\color{red}{r}} \binom{n+\color{red}{(n−r)}}{n}=2^{2n}$. Can someone please rectify my MathJax?
Why can I bring $2^{2n}$ on the RHS into the indexed variable (representing each term of the sum)? On the LHS, $n$ are the lower and upper bounds of summation. So $n$ isn't a Free Variable. I don't know if I'm using "Free Variable" correctly, but what I mean is that because if the RHS were $2^{2x}$, then I can bring $2^{2x}$ into the indexed variable because $x$ never appears in the capital-sigma notation.