How to show $\sum_{j=0}^{k} 2^{k-j}{k + j \choose j} = 4^k$

Question

I have a proof by induction going but I'm stuck. It goes as follows.

Let $$S_{k} = \sum_{j=0}^{k} 2^{k-j}{k + j \choose j}.$$

Then, $$S_{0} = \sum_{j=0}^{0} 2^{-j}{j \choose j} = 4^0$$

And,

$$S_{k+1} = \sum_{j=0}^{k+1} 2^{k+1-j}{(k+1)+j \choose j}$$

$$= 2 \sum_{j=0}^{k+1} 2^{k-j} \biggl( {k+j \choose j} + {k+j \choose j-1} \biggl)$$

$$= 2 \biggl( \sum_{j=0}^{k+1} 2^{k-j} {k+j \choose j} + \sum_{j=0}^{k+1} 2^{k-j} {k+j \choose j-1} \biggl)$$

$$= 2 \biggl( \sum_{j=0}^{k} 2^{k-j} + \frac{1}{2} {2k+1 \choose k+1} + \sum_{j=0}^{k} 2^{k-j} {k+j \choose j-1} + \frac{1}{2} {2k+1 \choose k} \biggl)$$

$$= 2S_{k} + {2k+1 \choose k+1} + {2k+1 \choose k} + 2\sum_{j=0}^{k} 2^{k-j} {k+j \choose j-1}$$

$$= 2S_{k} + 2{2k+1 \choose k} + 2\sum_{j=0}^{k} 2^{k-j} {k+j \choose j-1}$$

Where can I go from here? I'm trying to find a way to get another $2S_k$, making the right side equal $4S_k$, and I think that has to be done by combining the $2{2k+1 \choose k}$ with the summation, but I'm not sure how to do that. Thanks in advance.

Answer 1

I prefer a combinatorial solution. Consider the set $[2k + 1] = \{\,1, 2, \ldots, 2k + 1\,\}$. Let's count the number $x$ of subsets with cardinality at least $k + 1$.

On one hand there is one-to-one mapping between subsets with carinality at least $k + 1$ and subsets with cardinality at most $k$. And the total number of subsets of $[2k + 1]$ is $2^{2k + 1}$. Therefore $x = \frac{2^{2k + 1}}{2} = 2^{2k} = 4^k$.

On the other hand if subset contains exactly $k$ elements from $[k + j]$ and element $k + j + 1$ then there are $\binom{k + j}{k} = \binom{k + j}{j}$ ways to select first $k$ elements and $2^{2k + 1 - k - j - 1} = 2^{k - j}$ ways to select remaining elements. The total number of subsets is $x = \sum_{j = 0}^{k} 2^{k - j} \binom{k + j}{j}$. Q. E. D.

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If you need only formal proof by induction then congrats, you have a half. Let complete it. You have

$$x_{k + 1} = 2x_{k} + 2\binom{2k + 1}{k} + 2\sum_{j = 0}^{k} 2^{k - j} \binom{k + j}{j - 1} =\\= 2x_k + 2\binom{2k + 1}{k} + 2\sum_{j = 1}^{k} 2^{k - j} \binom{k + j}{j - 1} =\\= 2x_k + 2\binom{2k + 1}{k} + 2\sum_{\ell = 0}^{k - 1} 2^{k - \ell - 1} \binom{k + \ell + 1}{\ell} =\\= 2x_k + 2\binom{2k + 1}{k} + \sum_{\ell = 0}^{k - 1} 2^{k - \ell} \binom{k + \ell + 1}{\ell} = \\ = 2x_k + \binom{2k + 1}{k} + \sum_{\ell = 0}^{k} 2^{k - \ell} \binom{k + 1 + \ell}{\ell} =\\= 2x_k + \frac12\binom{2k + 2}{k + 1} + \frac12\sum_{\ell = 0}^{k} 2^{k + 1 - \ell} \binom{k + 1 + \ell}{\ell} = \\ = 2x_k + \frac12\sum_{\ell = 0}^{k + 1} 2^{k + 1 - \ell} \binom{k + 1 + \ell}{\ell} =\\= 2x_k + \frac12 x_{k + 1}.$$ Therefore $x_{k + 1} = 4x_k = 4^{k + 1}$.