Question

If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\dots \dots +10(11{)}^9=k(10{)}^9$, then k is equal to

• A. $\frac{121}{10}$
• B. $\frac{441}{100}$
• C. $100$
• D. $110$

Answer 1

The correct option is C $100$

Given
$k{.10}^9={10}^9+2\left(11{)}^1\right(10{)}^8+3\left(11{)}^2\right(10{)}^7+\dots \dots +10(11{)}^9$
$\Rightarrow k=1+2\left(\frac{11}{10}\right)+3{\left(\frac{11}{10}\right)}^2+\dots \dots +10{\left(\frac{11}{10}\right)}^9\dots \dots \left(i\right)$
$\left(\frac{11}{10}\right)k=1\left(\frac{11}{10}\right)+2{\left(\frac{11}{10}\right)}^9+10{\left(\frac{11}{10}\right)}^{10}\dots \dots \left(ii\right)$
On subtracting Eq. (ii) from Eq. (i), we get
$k(1-\frac{11}{10})=1+\frac{11}{10}+{\left(\frac{11}{10}\right)}^2+\dots \dots +{\left(\frac{11}{10}\right)}^9-10{\left(\frac{11}{10}\right)}^{10}$
$\Rightarrow k\left(\frac{10-11}{10}\right)=\frac{1[{\left(\frac{11}{10}\right)}^{10}-1]}{(\frac{11}{10}-1)}-10{\left(\frac{11}{10}\right)}^{10}$
$[\because InGP,sumofnterms=\frac{a(r^n-1)}{r-1},whenr>1]$
$\Rightarrow -k=10[10{\left(\frac{11}{10}\right)}^{10}-10-10{\left(\frac{11}{10}\right)}^{10}]$
$\therefore k=100$