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I was asked to show that

$\sum^{n}_{k=0}\binom{n+k}{k}2^{-k} = 2^n$ using pascal identity. I don't know how to do it. Whenever I was trying to use $\binom{n+k+1}{k} = \binom{n+k}{k}+\binom{n+k}{k-1}$, my index is always off by 1.

metricspace
  • 851
  • Write your proof out term by term for n = 1 (proving the induction step to n = 1 from n = 0) to debug it. – Count Iblis Oct 18 '16 at 02:08
  • The earlier question didn’t specify any particular method of proof, but the accepted answer is by induction using Pascal’s identity, and zyx gives a nice combinatorial proof. – Brian M. Scott Oct 18 '16 at 13:00

2 Answers2

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} a_{n} & \equiv \sum_{k = 0}^{n}{n + k \choose k}2^{-k} = \sum_{k = 0}^{n}{n + k - 1 \choose k}2^{-k} + \sum_{k = 1}^{n}{n + k -1 \choose k - 1}2^{-k} \\[5mm] &= \sum_{k = 0}^{n}{n + k - 1 \choose k}2^{-k} + \sum_{k = 0}^{n - 1}{n + k \choose k}2^{-k - 1} \\[5mm] &= \bracks{\underbrace{\sum_{k = 0}^{n - 1}{n + k - 1 \choose k}2^{-k}} _{a_{n - 1}} + {2n - 1 \choose n}2^{-n}} + {1 \over 2}\bracks{\underbrace{\sum_{k = 0}^{n}{n + k \choose k}2^{-k}} _{\ds{a_{n}}} - {2n \choose n}2^{-n}} \end{align}

\begin{align} a_{n} & = 2a_{n - 1} + {2n - 1 \choose n}2^{-n + 1} - {2n \choose n}2^{-n} \\[5mm] & = 2a_{n - 1} + {2n - 1 \choose n}2^{-n + 1} - \bracks{{2n - 1 \choose n} + {2n - 1 \choose n - 1}}2^{-n} \\[5mm] & = 2a_{n - 1} + {2n - 1 \choose n}2^{-n} - {2n - 1 \choose n - 1}2^{-n} = 2a_{n - 1} + {2n - 1 \choose n}2^{-n} - {2n - 1 \choose n}2^{-n} \\[5mm] \implies & \bbx{\ds{a_{n} = 2a_{n - 1}}} \end{align}
$$ a_{n} \equiv \sum_{k = 0}^{n}{n + k \choose k}2^{-k} = 2a_{n - 1} = 2^{2}a_{n - 2} = 2^{3}a_{n - 3} = \cdots = 2^{n}a_{0} = \bbx{\ds{2^{n}}}\quad\mbox{because}\quad a_{0} = 1 $$
Felix Marin
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1

Pascal distribution

Let $f$ be the probability mass function of the Pascal distribution with $r=n+1$ and $p=1-p=\frac 1 2$, $$f(k;n+1,\frac 1 2)=\Pr(X=k)={\binom {k+n}{k}}p^{k}(1-p)^{r}\quad {\text{for }}k \in \Bbb N$$then $$ \Pr(X\leq n)= \sum^{n}_{k=0}\binom{n+k}{k}(\frac 1 2)^k(\frac 1 2)^{n+1} $$ Since $\Pr(X\leq n)=I_{\frac 1 2}(n+1,n+1)$, and $$ I_{\frac 1 2}(n+1,n+1)={\frac {\mathrm {B} ({\frac 1 2};\,n+1,n+1)}{\mathrm {B} (n+1,n+1)}}=\frac 1 2 $$ we have $$\Pr(X\leq n)=\sum^{n}_{k=0}\binom{n+k}{k}(\frac 1 2)^k(\frac 1 2)^{n+1}=\frac 1 2$$ thus, $$\sum^{n}_{k=0}\binom{n+k}{k}(\frac 1 2)^k=2^{n+1}\frac 1 2=2^n$$

Aforest
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