Showing $ \frac{2^j(2k-j)!}{(k-j)!}=\sum_{l=j}^k\frac{2^ll(2k-l-1)!}{(k-l)!} $

Question

For any given $j,k\in\mathbb{N}$, with $j\leq k$, how is it possible to prove the following equivalence? $$ \frac{2^j(2k-j)!}{(k-j)!}=\sum_{l=j}^k\frac{2^ll(2k-l-1)!}{(k-l)!} $$ My attempt is: $$ \frac{2^j(2k-j)!}{(k-j)!}=\sum_{l=j}^k\frac{2^ll(2k-l-1)!}{(k-l)!}\\ 2^j\binom{2k-j}{k-j}k!=\sum_{l=j}^k\binom{2k-l-1}{k-l}l2^l(k-1)!\\\binom{2k-j}{k-j}k=\sum_{l=j}^kl2^{l-j} \binom{2k-l-1}{k-l} $$ and then I don't know how to continue. Maybe there is some binomial formula for this sum, but I don't know.

Answer 1

Here is a generating function approach that generalises @MarkoRiedel's answer to this problem. We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. This way we can write for instance \begin{align*} [z^k](1+z)^n = \binom{n}{k}\tag{1} \end{align*}

We show the following identity is valid \begin{align*} \color{blue}{\sum_{l=j}^k\binom{2k-l-1}{k-l}l\,2^{l-j}=k\binom{2k-j}{k-j}}\tag{2} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{l=j}^k}&\color{blue}{\binom{2k-l-1}{k-l}l\,2^{l-j}}\\ &=\sum_{l=j}^k[z^{k-l}](1+z)^{2k-l-1}l\,2^{l-j}\tag{3}\\ &=[z^k](1+z)^{2k-1}\sum_{l=j}^\infty\left(\frac{z}{1+z}\right)^ll\,2^{l-j}\tag{4}\\ &=[z^k](1+z)^{2k-1}\sum_{l=0}^\infty\left(\frac{z}{1+z}\right)^{l+j}(l+j)2^l\tag{5}\\ &=[z^{k-j}](1+z)^{2k-j-1}\sum_{l=0}^{\infty}\left(\frac{2z}{1+z}\right)^l(l+j)\\ &=[z^{k-j}](1+z)^{2k-j-1}\left(\frac{2z(1+z)}{(1-z)^2}+j\frac{1+z}{1-z}\right)\tag{6}\\ &=2[z^{k-j}](1+z)^{2k-j}\frac{z}{(1-z)^2}+j[z^{k-j}](1+z)^{2k-j}\frac{1}{1-z}\\ &=2\sum_{m=0}^{k-j}\binom{2k-j}{m}(k-j-m)+j\sum_{m=0}^{k-j}\binom{2k-j}{m}\tag{7}\\ &=\sum_{m=0}^{k-j}\binom{2k-j}{2k-j-m}(2k-j-m)-\sum_{m=1}^{k-j}\binom{2k-j}{m}m\tag{8}\\ &=(2k-j)\sum_{m=0}^{k-j}\binom{2k-j-1}{2k-j-m-1}-(2k-j)\sum_{m=1}^{k-j}\binom{2k-j-1}{m-1}\tag{9}\\ &=(2k-j)\sum_{m=0}^{k-j}\binom{2k-j-1}{m}-(2k-j)\sum_{m=0}^{k-j-1}\binom{2k-j-1}{m}\\ &=(2k-j)\binom{2k-j-1}{k-j}=(2k-j)\binom{2k-j-1}{k-1}\\ &\,\,\color{blue}{=k\binom{2k-j}{k}} \end{align*} and the claim (2) follows.

Comment:

• In (3) we apply the coefficient of operator according to (1).

• In (4) we apply $[z^{p-q}]A(z)=[z^p]z^qA(z)$ and some rearrangements. We also set the upper limit to $\infty$ without changing anything thanks to $[z^k]$.

• in (5) we shift the index to start with $j=0$.

• In (6) we recall the binomial series expansions \begin{align*} \frac{z}{(1-z)^2}=\sum_{q=1}^{\infty}qz^q\qquad\qquad\frac{1}{1-z}=\sum_{q=0}^{\infty}z^q \end{align*}

• In (7) we select the coefficient of $z^{k-j}$ recalling the Cauchy product of series.

• In (8) we apply $\binom{p}{q}=\binom{p}{p-q}$ and rearrange the sums.

• In (9) we apply $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

Answer 2

Separate the last term of your sum as $$ \sum\limits_{\ell = j}^{k - 1} {\frac{{2^\ell \ell (2k - \ell - 1)!}}{{(k - \ell )!}}} + \frac{{2^k k(2k - k - 1)!}}{{(k - k)!}} = \sum\limits_{\ell = j}^{k - 1} {\frac{{2^\ell \ell (2k - \ell - 1)!}}{{(k - \ell )!}}} + 2^k k!, $$ and note the telescoping property $$ \frac{{2^\ell \ell (2k - \ell - 1)!}}{{(k - \ell )!}} = \frac{{2^\ell (2k - \ell )!}}{{(k - \ell )!}} - \frac{{2^{\ell + 1} (2k - (\ell + 1))!}}{{(k - (\ell + 1))!}}, $$ for $j\le \ell \le k-1$.

Answer 3

Using induction:

Let the RHS be $f(j)$, with $0\le j\le k$. We prove by induction that: $$P(j): f(j) = \color{red}{\frac{2^j(2k-j)!}{(k-j)!}}$$is true for $j \le k$.

For the base case $j=k$, note that the sum has only one summand, i.e, $$f(\color{blue}{k}) = \frac{2^{\color{blue}{k}}\color{blue}{k} (2k-\color{blue}{k}-1)!}{(k-\color{blue}{k})!} = \color{red}{\frac{2^k(2k-k)!}{(k-k)!}}$$ So, $P(k)$ is true.

Suppose $P(a+1)$ holds for $k-1\ge a$. We need to prove $P(a)$ is true. Note that: $$\begin{align}f(a) &= f(a+1) + \frac{2^aa(2k-a-1)!}{(k-a)!} \\ &= \color{red}{\frac{2^{a+1}(2k-a-1)!}{(k-a-1)!}} + \frac{2^aa(2k-a-1)!}{(k-a)!} \\ &= \frac{2^a(2k-a-1)!}{(k-a-1)!}\left(2+\frac{a}{k-a}\right) \\ &= \frac{2^a(2k-a-1)!}{(k-a-1)!}\cdot \frac{2k-a}{k-a} \\ &= \color{red}{\frac{2^a(2k-a)!}{(k-a)!}} \tag*{$\blacksquare$}\\ \end{align}$$