If ∑ (10^(r+1)-1/10^r). 11Cr+1 r ∈ to (r=0, 10) =α^(11)-11^(11)/10^(10), then α is equal to:

Question

If \(\sum_\limits{\mathrm{r}=0}^{10}\left(\frac{10^{\mathrm{r}+1}-1}{10^{\mathrm{r}}}\right) \cdot{ }^{11} \mathrm{C}_{\mathrm{r}+1}=\frac{\alpha^{11}-11^{11}}{10^{10}},\) then \(\alpha\) is equal to :

(1) 15

(2) 11

(3) 24

(4) 20  

Answer 1

Correct option is: (4) 20   

\(\sum_\limits{\mathrm{r}=0}^{10}\left(\frac{10^{\mathrm{r}-1}-1}{10^{r}}\right)^{11} \mathrm{C}_{\mathrm{r}+1}\)

\(=\sum_\limits{\mathrm{r}=0}^{10}\left(10-\frac{1}{10^{\mathrm{r}}}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1}\)

\(=10 \sum_\limits{\mathrm{r}=0}^{10}{ }^{11} \mathrm{C}_{\mathrm{r}+1}-10 \sum\left({ }^{11} \mathrm{C}_{\mathrm{r}+1}\left(\frac{1}{10}\right)^{\mathrm{r}+1}\right)\)

\(=10\left[{ }^{11} \mathrm{C}_{1}+{ }^{11} \mathrm{C}_{2}+\ldots . .+{ }^{11} \mathrm{C}_{11}\right]\)

\(-10\left[{ }^{11} \mathrm{C}_{1}\left(\frac{1}{10}\right)^{1}+{ }^{11} \mathrm{C}_{2}\left(\frac{1}{10}\right)^{2}+\ldots . .+{ }^{11} \mathrm{C}_{11}\left(\frac{1}{10}\right)^{11}\right]\)

\(=10\left[2^{11}-1\right]-10\left[\left(1+\frac{1}{10}\right)^{11}-1\right]\)

\(=10(2)^{11}-10-\frac{11^{11}}{10^{10}}+10\)

\(=\frac{(20)^{11}-11^{11}}{10^{10}}\)

\(\therefore\ \alpha=20\)