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I'm struggling to understand the following derivation where $n$ is a positive integer.

$$ \sum_{\ell=0}^n {n \choose \ell} 2^\ell \log 2^\ell = n \sum_{\ell=0}^{n-1} {n-1 \choose \ell} 2^{\ell+1}. $$ Splitting off the last term of the sum I can see where the $n$ factor comes from but I'm not sure why this changes the binomial coefficient.

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  • $\begingroup$ I wonder if there is something missing, try $n=1$ or $n=2$. $\endgroup$ – user201043 Dec 14 '14 at 8:02
  • $\begingroup$ Are you using the base $2$ logarithm? $\endgroup$ – Robert Israel Dec 14 '14 at 8:12
  • $\begingroup$ @RobertIsrael: Yes $\endgroup$ – werderman Dec 14 '14 at 8:12
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I replace with $l$ with $k$ for better readability

$$\log(2^k)=k\cdot\log2$$

and $$k\binom nk=k\cdot\frac{n!}{(n-k)!\cdot k!}=n\binom{n-1}{k-1}$$

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