Question

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

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

Answer 1

Answer: (C)

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