Question

The value of $({10}^{11}-{10}^9-2\times 11\times {10}^8-3\times {11}^2\times {10}^7-.....-(10)\times {11}^9)$ is equal to

• A. $0$
• B. ${11}^{10}$
• C. ${11}^{11}$
• D. ${10}^{11}$

Answer 1

The correct option is A $0$

$T={10}^{11}-\{{10}^9+2\times 11\times {10}^8+........+(10)\times {11}^9\}$

$={10}^{11}-{\sum }_{r=1}^{10}r.{10}^{10-r}.{11}^{r-1}$

$={10}^{11}-\Sigma r.\frac{{10}^{10}}{{10}^r}.\frac{{11}^r}{11}$

$T={10}^{11}-\frac{{10}^{10}}{11}{\sum }_{r=1}^{10}r.{\left(\frac{11}{10}\right)}^r$

${\sum }_{r=1}^{10}r.{\left(\frac{11}{10}\right)}^r=S$

$S=\left(\frac{11}{10}\right)+2{\left(\frac{11}{10}\right)}^2+3{\left(\frac{11}{10}\right)}^3+....10{\left(\frac{11}{10}\right)}^{10}$

$\left(\frac{11}{10}\right)S={\left(\frac{11}{10}\right)}^2+2{\left(\frac{11}{10}\right)}^3+.....9\left(\frac{11}{10}\right)+10{\left(\frac{11}{10}\right)}^{11}$

$=(1-\frac{11}{10})S=\frac{11}{10}+{\left(\frac{11}{10}\right)}^2+{\left(\frac{11}{10}\right)}^3+.....{\left(\frac{11}{10}\right)}^{10}-\left(\frac{{11}^{11}}{{10}^{10}}\right)$

$\Rightarrow \frac{-S}{10}=\frac{\frac{11}{10}(1-{\left(\frac{11}{10}\right)}^{10})}{1-\frac{11}{10}}-\left(\frac{{11}^{11}}{{10}^{10}}\right)$

$\Rightarrow S=\left(\frac{{11}^{11}}{{10}^{10}}\right)\times 10-11\times 10\{1-{\left(\frac{11}{10}\right)}^{10}\}$

$\Rightarrow S=\frac{{11}^{11}}{{10}^9}+110-\frac{{11}^{11}}{{10}^9}=110$

$\Rightarrow T={10}^{11}-\frac{{10}^{10}}{11}\times 10\times 10$

$\Rightarrow {10}^{11}-{10}^{11}=0.$

Hence the answer is $0.$