What is tried was : $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{Note that inside summation is }\quad\left. \sum_{k = 0}^{r}\pars{-1}^{k}\pars{k + 1}\pars{k + 2}{2021 \choose r - k} \right\vert_{\ 2021\ >\ 3} = \left.\partiald[2]{}{x}\sum_{k = 0}^{r}{2021 \choose r - k}x^{k + 2} \right\vert_{\ x\ =\ -1} \end{align} , now how to simplfiy the summation part which needs to be double differentiated or there is some other method to solve for that summation value too ?
-
I don't think your Latex rendered properly, please refer the MathJax reference. – person May 02 '21 at 01:22
-
Done @person... – Orion_Pax May 02 '21 at 01:25
-
There isn’t even a good closed form for $$\sum_{k=0}^r\binom{n}{r-k}=\sum_{k=0}^r\binom nk$$ so there probably isn’t a closed form for $$\sum_{k=0}^r\binom n{r-k}x^{k+2}$$ – Thomas Andrews May 02 '21 at 01:27
-
Yes i also think same but i meant the summation of inside part has a closed form of 2*n-3Cr , i am not sure of the closed form for the summation which needed to be diff twice – Orion_Pax May 02 '21 at 02:13
-
Anything good happen if you reverse the order of summation? – Gerry Myerson May 02 '21 at 06:42
-
Why not if its leading to something gud by adding them both ? @GerryMyerson and btw though what will be expression looks like after swapping the order? – Orion_Pax May 02 '21 at 19:57
-
$\sum_{r=0}^{2019}\sum_{k=0}^r$ becomes $\sum_{k=0}^{2019}\sum_{r=k}^{2019}$. – Gerry Myerson May 02 '21 at 23:01
2 Answers
In seeking to evaluate
$$\sum_{r=0}^n \sum_{k=0}^r (-1)^k (k+1) (k+2) {n+2\choose r-k}$$
we write
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2} \sum_{k=0}^r {k+2\choose 2} (-1)^k z^k.$$
The coefficient extractor enforces the range of the inner sum:
$$2 \sum_{r=0}^n [z^r] (1+z)^{n+2} \sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k \\ = 2 \sum_{r=0}^n [z^r] (1+z)^{n+2} \frac{1}{(1+z)^3} = 2 \sum_{r=0}^n [z^r] (1+z)^{n-1} \\ = 2 \sum_{r=0}^n {n-1\choose r} = 2 \times 2^{n-1} = 2^n$$
where we have used $n\ge 1,$ for $n=0$ we find $2[z^0] (1+z)^{-1} = 2.$
Addendum. We may also change the order of summation as asked in the comments to get
$$2 \sum_{k=0}^n {k+2\choose 2} (-1)^k \sum_{r=k}^n {n+2\choose r-k} = 2 \sum_{k=0}^n {k+2\choose 2} (-1)^k \sum_{r=0}^{n-k} {n+2\choose r} \\ = 2 \sum_{k=0}^n {k+2\choose 2} (-1)^k [z^{n-k}] \frac{1}{1-z} (1+z)^{n+2} \\ = 2 [z^n] \frac{1}{1-z} (1+z)^{n+2} \sum_{k=0}^n {k+2\choose 2} (-1)^k z^k.$$
We once more have a coefficient extractor enforcing the upper limit of the sum and we find
$$2 [z^n] \frac{1}{1-z} (1+z)^{n+2} \sum_{k\ge 0} {k+2\choose 2} (-1)^k z^k = 2 [z^n] \frac{1}{1-z} (1+z)^{n+2} \frac{1}{(1+z)^3} \\ = 2 [z^n] \frac{1}{1-z} (1+z)^{n-1} = 2 \sum_{r=0}^{n-1} {n-1\choose r} = 2 \times 2^{n-1} = 2^n.$$
Here we see that keeping the original order of the two summations was the better method moreover no new features appeared on changing the order.
- 61,317
-
Sir how u arrived at second thing from first pls explain in detail as such most terms r dependemt on k and how u made it to be of something coefficient of z^r : $$\sum_{r=0}^n \sum_{k=0}^r (-1)^k (k+1) (k+2) {n+2\choose r-k}$$ = $$2 \sum_{r=0}^n [z^r] (1+z)^{n+2} \sum_{k=0}^r {k+2\choose 2} (-1)^k z^k.$$ – Orion_Pax May 02 '21 at 19:55
-
${n+2\choose r-k} = [z^{r-k}] (1+z)^{n+2} = [z^r] z^k (1+z)^{n+2}.$ – Marko Riedel May 02 '21 at 20:26
-
Well explained Sir got your pt ,btw Sir what u think of this thing the solution writer wrote do u think its correct as he got n-3Cr instead of yours n-1Cr ?https://math.stackexchange.com/a/2225725/922054 – Orion_Pax May 02 '21 at 23:29
-
-
Got the pt , ty Sir , also do u think Sir that changing the order of summation will lead to anything good ? – Orion_Pax May 02 '21 at 23:37
-
-
-
Sir one last thing how u got that inside summatiom to be equal to 1/(1+z)^3 , as such shouldnt it depend on r as such ? As summation is from k=0 to r isnt ? I didnt get the meaning of "coefficient extrafactor enforces the inner range " maybe that is related to this – Orion_Pax May 03 '21 at 13:17
-
-
Wonderful Sir , is this method very famous and is there a name to it and r there any more such problems here in stack exchange u remember ? Where coefficient we take out and also in the inside function . – Orion_Pax May 03 '21 at 16:38
-
This is the Egorychev method in formal power series. It may be extended to complex variables but that was not necessary here. – Marko Riedel May 03 '21 at 18:14
-
Hint
I think that this is a red herring.
Consider instead $$S_n=\sum_{r= 0}^{n-2}\sum_{k=0}^{r}(-1)^k\,(k+1)\,(k+2)\binom{n}{r-k}$$ and compute the very first values of $S_n$. You should find avery clear pattern.
- 260,315
-
But this pattern thing , its not a accurate method isnt ? As such if we do this in this case then for all binomial problems we can do stuff and check out a pattern ig isnt Sir? – Orion_Pax May 02 '21 at 06:38
-
@Orion_Pax. I totally agree with you that this does not prove anything. This was just to give you an idea of what you are looking for. Cheers :-) – Claude Leibovici May 02 '21 at 06:41
-
1Finding a pattern is a first step, Orion. But it tells you what you're looking for. Then, you can try to prove the pattern holds, say, by induction. – Gerry Myerson May 02 '21 at 06:41