2022-07-03

Having trouble with proving this binomial identity. I have taken it as far as I can go.
I have broken down this equation into factorials, but I'm unsure of where to go from here. This may not even be the right approach to solve this binomial transform. Any help would be appreciated.
Binomial transform identity:
$\begin{array}{rl}& \sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{n+jk}{jk}\right)={j}^{n}\\ & \sum _{k=0}^{n}\left(-1{\right)}^{n-k}\frac{\left(jk+1\right)\left(jk+2\right)\dots \left(jk+n\right)}{\left(n-k\right)!\phantom{\rule{thinmathspace}{0ex}}k!}={j}^{n}\end{array}$

Kiley Hunter

Expert

Step 1
In trying to prove
$\sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{n+jk}{jk}\right)={j}^{n}$
$\sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{n+jk}{n}\right)\phantom{\rule{0ex}{0ex}}=\left[{z}^{n}\right]\left(1+z{\right)}^{n}\sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(1+z{\right)}^{jk}\phantom{\rule{0ex}{0ex}}=\left[{z}^{n}\right]\left(1+z{\right)}^{n}\left(\left(1+z{\right)}^{j}-1{\right)}^{n}.$.
Step 2
Now $\left(\left(1+z{\right)}^{j}-1{\right)}^{n}=\left(jz+\cdots {\right)}^{n}$ so the only term that contributes to the coefficient extractor in z is ${j}^{n}{z}^{n}.$. (The n factors contribute at least z so for the power to be less than or equal to n we have to choose z from each factor. As soon as we choose just one factor to a power at least two the term produces a power that is larger than n and does not contribute to the coefficient extractor.) We get $\left[{z}^{n}\right]\left(1+z{\right)}^{n}{j}^{n}{z}^{n}={j}^{n}$ as claimed.

Alissa Hancock

Expert

Step 1
Recall that, if p(x) is a polynomial of degree n,
$\sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right)p\left(x+k\right)$ is n! times the leading coefficient of p(x) (this may be proven by induction). Now, select
$p\left(x\right)=\frac{\left(jx+n\right)\left(jx+n-1\right)\cdots \left(jx+1\right)}{n!},$,
which is a polynomial of degree n with leading coefficient ${j}^{n}/n!$. For integer k,
$p\left(k\right)=\frac{\left(jk+n\right)\left(jk+n-1\right)\cdots \left(jk+1\right)}{n!}=\left(\genfrac{}{}{0}{}{jk+n}{n}\right)=\left(\genfrac{}{}{0}{}{n+jk}{jk}\right).$.
Step 2
So, $\sum _{k=0}^{n}\left(-1{\right)}^{n-k}\left(\genfrac{}{}{0}{}{n}{k}\right)\left(\genfrac{}{}{0}{}{n+jk}{jk}\right)$ is n! times the leading coefficient of p(x), i.e. ${j}^{n}$.

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