If \(\sum _{\mathrm{r}=0}^{10}\left(\frac{{10}^{\mathrm{r}+1}-1}{{10}^{\mathrm{r}}}\right)\cdot {}^{11}\mathrm{C}_{\math…

Explanation

Let \(k=r+1\). When \(r=0\), \(k=1\); when \(r=10\), \(k=11\). The sum becomes:

\(\sum _{k=1}^{11}(\frac{11}{k})\)

Using the binomial theorem, \({\sum }_{k=0}^{11}(\frac{11}{k})={2}^{11}\). Thus:

\(\sum _{k=1}^{11}(\frac{11}{k})={2}^{11}−(\frac{11}{0})={2}^{11}−1\)

Multiplying by 10:

\(10\sum _{r=0}^{10}(\frac{11}{r+1})=10({2}^{11}−1)\)

Step 2: Evaluate the second sum \({\sum }_{r=0}^{10}\frac{(\frac{11}{r+1})}{{10}^{r}}\)

Again, let \(k=r+1\), so \(r=k−1\). The sum becomes:

\(\sum _{k=1}^{11}\frac{(\frac{11}{k})}{{10}^{k−1}}=10\sum _{k=1}^{11}\frac{(\frac{11}{k})}{{10}^{k}}\)

Using the binomial theorem for \({(1+\frac{1}{10})}^{11}\):

\({(1+\frac{1}{10})}^{11}=\sum _{k=0}^{11}(\frac{11}{k}){(\frac{1}{10})}^{k}\text{  }⟹\text{  }\sum _{k=0}^{11}\frac{(\frac{11}{k})}{{10}^{k}}={(\frac{11}{10})}^{11}\)

Subtracting the \(k=0\) term:

\(\sum _{k=1}^{11}\frac{(\frac{11}{k})}{{10}^{k}}={(\frac{11}{10})}^{11}−1\)

Multiplying by 10:

\(10\sum _{k=1}^{11}\frac{(\frac{11}{k})}{{10}^{k}}=10(\frac{{11}^{11}}{{10}^{11}}−1)=\frac{{11}^{11}}{{10}^{10}}−10\)

Step 3: Combine the two sums

Substitute the results back into the original expression:

\(10({2}^{11}−1)−(\frac{{11}^{11}}{{10}^{10}}−10)=10⋅{2}^{11}−10−\frac{{11}^{11}}{{10}^{10}}+10=10⋅{2}^{11}−\frac{{11}^{11}}{{10}^{10}}\)

Notice that \(10⋅{2}^{11}=\frac{(20{)}^{11}}{{10}^{10}}\) (since \({20}^{11}=(2⋅10{)}^{11}={2}^{11}⋅{10}^{11}\), so dividing by \({10}^{10}\) gives \({2}^{11}⋅10\)). Thus:

\(10⋅{2}^{11}−\frac{{11}^{11}}{{10}^{10}}=\frac{{20}^{11}−{11}^{11}}{{10}^{10}}\)

Comparing with the given form \(\frac{{\alpha }^{11}−{11}^{11}}{{10}^{10}}\), we see \(\alpha =20\).