If ( 10 ) 9 + 2 ( 11 ) 1 ( 10 ) 8 + 3 ( 11 ) 2 ( 10 ) 7 + ...... + 10 ( 11 ) 9   = k ( 10 ) 9 , then k…

Let,$S={\left(10\right)}^9+2{\left(11\right)}^1 {\left(10\right)}^8+3{\left(11\right)}^2 {\left(10\right)}^7+\mathrm{......}+10{\left(11\right)}^9 =k{\left(10\right)}^9$

$S={\left(10\right)}^9\left\{1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+.....+10{\left(\frac{11}{10}\right)}^9\right\}$

So,$k=1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+.....+10{\left(\frac{11}{10}\right)}^9$

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

Subtracting

$-\left(\frac{k}{10}\right)=1+\left(\frac{11}{10}\right)+{\left(\frac{11}{10}\right)}^2+.\dots \dots +{\left(\frac{11}{10}\right)}^9-10{\left(\frac{11}{10}\right)}^{10}$
Here,

First term $=1$

Common ratio$=\frac{11}{10}$

Number of terms$=10$

$\therefore \frac{k}{10}=10{\left(\frac{11}{10}\right)}^{10}-\left[\frac{1\left\{{\left(\frac{11}{10}\right)}^{10}-1\right\}}{\left\{\frac{11}{10}-1\right\}}\right]$

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

$k=\left[100{\left(\frac{11}{10}\right)}^{10}-100{\left(\frac{11}{10}\right)}^{10}+100\right]$
$k=100$