Prove that: $ \sum_{r=0}^n (-1)^r (^nCr_) \frac{1 + r\log_e10}{(1 + log_e10^n)^r} = 0 $ Prove that:
$$ \sum_{r=0}^n (-1)^r (^nCr_) \frac{1 + r\log_e10}{(1 + log_e10^n)^r} = 0 $$
Attempt:
$$ (1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n $$
$$ (1 + x)^n = \sum_{r=0}^nC_r(x^r)$$
$$ (1 + \frac{-1}{1 + log_e10^n})^n = \sum_{r=0}^n C_r(-1)^n\frac{1}{(1 + log_e10^n)^r}$$
I am unable to figure out how to resolve the numerator part in the question $ (1 + rlog_e10) $ into the sum. Clearly, the $ r $ here is not a power (except inside the logarithm).
 A: Two aspects:


*

*The claim is valid for positive integer $n$, but not for $n=0$ where evaluation of the left-hand side gives $1$.

*When looking at the formula we might assume that properties of the logarithm are essential, especially the rule $\log z^n=n\log z$, but neither the base nor the argument seem to be relevant for the claim. So, we can show a somewhat more general formula.

We obtain for integral $n\geq 0$ and $z>0$
\begin{align*}
\color{blue}{\sum_{r=0}^n}&\color{blue}{(-1)^r\binom{n}{r}\frac{1+r\log z}{(1+\log z^n)^r}}\\
&=\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{\left(1+\log z^n\right)^r}
+\sum_{r=1}^n(-1)^r\binom{n-1}{r-1}\frac{n\log z}{\left(1+\log z^n\right)^r}\tag{1}\\
&=1+\sum_{r=1}^{n-1}(-1)^r\left(\binom{n-1}{r}+\binom{n-1}{r-1}\right)\frac{1}{\left(1+\log z^n\right)^r}\\
&\qquad+(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}\tag{2}\\
&=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r}
-\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r+1}}\\
&\qquad+(-1)^n\frac{1}{(1+\log z^n)^r}-\sum_{r=0}^{n-1}(-1)^r\binom{n-1}{r}\frac{\log z^n}{\left(1+\log z^n\right)^{r+1}}\tag{3}\\
&=1+\sum_{r=1}^{n-1}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^r}+(-1)^n\frac{1}{(1+\log z^n)^r}\\
&\qquad-\sum_{r=0}^{n-2}(-1)^r\binom{n-1}{r}\frac{1}{\left(1+\log z^n\right)^{r}}+(-1)^n\frac{\log z^n}{(1+\log z^n)^n}\tag{4}\\
&=1+(-1)^{n-1}\frac{1}{(1+\log z^n)^{n-1}}+(-1)^n\frac{1}{(1+\log z^n)^{n-1}}\\
&\qquad-1+(-1)^n\frac{\log z}{(1+\log z^n)^n}\tag{5}\\
&\,\,\color{blue}{=0}\tag{6}
\end{align*}
  and the (generalised) claim follows.

Comment:


*

*In (1) we split the sum according to the terms in the numerator and use the binomial identity $q\binom{p}{q}=p\binom{p-1}{q-1}$.

*In (2) we separate the first and last term of the left sum and apply the binomial identity
$\binom{p}{q}=\binom{p-1}{q}+\binom{p-1}{q-1}$. We also shift the index of the right sum to start with $r=0$.

*In (3) we split the left sum and shift the right part of it to start with $r=0$.

*In (4) we separate the last term $r=n-1$ of the right-most sum and merge the second and the right-most sum. This way we can also cancel $1+\log z^n$.

*In (5) we cancel the terms of the sums with $1\leq r\leq n-2$.

*In (6) we observe that all other terms cancel as well.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[#ffd,10px]{\ds{\sum_{r = 0}^{n}\pars{-1}^{r}\pars{^{n}Cr}\,{1 + r\log_{\expo{}}\pars{10} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{r}}}}
\\[5mm] = &
\sum_{r = 0}^{n}{n \choose r}\bracks{-\,{1 \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{r}
 +
\left.\log_{\expo{}}\pars{10}\,\partiald{}{x}
\sum_{r = 0}^{n}{n \choose r}
\bracks{-\,{x \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{r}
\,\right\vert_{\ x\ =\ 1}
\\[5mm] = &\
\bracks{1 - {1 \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{n} +
\left.\log_{\expo{}}\pars{10}\,\partiald{}{x}\bracks{1 - {x \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{n}\,\right\vert_{\ x\ =\ 1}
\\[5mm] = &\
{\log_{\expo{}}^{n}\pars{10^{n}} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} +
\left.\log_{\expo{}}\pars{10}\,n\bracks{1 - {x \over 1 + \log_{\expo{}}\pars{10^{n}}}}^{n - 1}\bracks{-\,{1 \over 1 + \log_{\expo{}}\pars{10^{n}}}}\,\right\vert_{\ x\ =\ 1}
\\[5mm] = &\
{\log_{\expo{}}^{n}\pars{10^{n}} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} -
n\,{\log_{\expo{}}\pars{10}\log_{\expo{}}^{n - 1}\pars{10^{n}} \over
\bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}}
\\[5mm] = &\
{n^{n}\log_{\expo{}}^{n}\pars{10} \over \bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} -
n\,{\log_{\expo{}}\pars{10}\bracks{n^{n - 1}\log_{\expo{}}^{n - 1}\pars{10}} \over
\bracks{1 + \log_{\expo{}}\pars{10^{n}}}^{n}} = \bbx{0}
\end{align}
