Discussion Forum

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

Answer:

Explanation:

$k.10^{9}=10^{9}+2(11)^{1}(10)^{8}+3((11)^{2}(10)^{7}+....+10(11)^{9}$

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

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

  On subtracting Eq,(ii) from Eq.(i) , we get

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

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

[  $\because$    In GP sum of n terms =  $\frac{a(r^{n}-1)}{r-1}$  when r>1]

  $\Rightarrow$     -k

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

                 k=100