Question

# If ${{\left( 10 \right)}^{9}}+2\left( 11 \right){{\left( 10 \right)}^{8}}+3{{\left( 11 \right)}^{2}}{{\left( 10 \right)}^{7}}+....+10{{\left( 11 \right)}^{9}}=k{{\left( 10 \right)}^{9}}$, then k is equal to(a) $\dfrac{121}{10}$(b) $\dfrac{441}{100}$(c) 100(d) 110

Hint: First, before proceeding for this, we must suppose the given sum of series as S. Then, we need to make it somewhat simple to get the arithmetic progression from it to apply the formula for the sum of series. Then, we can see clearly from the above expression that all the terms are in geometric progression(GP) except the last term and by applying the formula for the sum of GP excluding the last term $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ where a is the first term and r is the common ratio to get the final answer.

In this question, we are supposed to find the value of k for the given sum of series as ${{\left( 10 \right)}^{9}}+2\left( 11 \right){{\left( 10 \right)}^{8}}+3{{\left( 11 \right)}^{2}}{{\left( 10 \right)}^{7}}+....+10{{\left( 11 \right)}^{9}}=k{{\left( 10 \right)}^{9}}$.
So, before proceeding for this, we must suppose the given sum of series as S so we get:
$S={{\left( 10 \right)}^{9}}+2\left( 11 \right){{\left( 10 \right)}^{8}}+3{{\left( 11 \right)}^{2}}{{\left( 10 \right)}^{7}}+....+10{{\left( 11 \right)}^{9}}....\left( i \right)$
Now, we need to make it somewhat simple to get the arithmetic progression from it to apply the formula for the sum of series.
So, by multiplying both sides by $\dfrac{11}{10}$, we get:
$\dfrac{11}{10}S=11{{\left( 10 \right)}^{8}}+2{{\left( 11 \right)}^{2}}{{\left( 10 \right)}^{7}}+3{{\left( 11 \right)}^{3}}{{\left( 10 \right)}^{6}}+....+{{\left( 11 \right)}^{10}}....\left( ii \right)$
Now, by subtracting the equation (i) from equation (ii), we get:
$\dfrac{11}{10}S-S=\left( 11{{\left( 10 \right)}^{8}}+2{{\left( 11 \right)}^{2}}{{\left( 10 \right)}^{7}}+3{{\left( 11 \right)}^{3}}{{\left( 10 \right)}^{6}}+....+{{\left( 11 \right)}^{10}} \right)-\left( {{\left( 10 \right)}^{9}}+2\left( 11 \right){{\left( 10 \right)}^{8}}+3{{\left( 11 \right)}^{2}}{{\left( 10 \right)}^{7}}+....+10{{\left( 11 \right)}^{9}} \right)$Then, by solving the above expression, we get:
$\dfrac{S}{10}={{10}^{9}}+11\times {{10}^{8}}+{{11}^{2}}\times {{10}^{7}}+....+{{11}^{8}}\times 10+{{11}^{9}}-{{11}^{10}}$
Now, we can see clearly from the above expression that all the terms are in geometric progression(GP) except the last term.
Then, by applying the formula for the sum of GP excluding the last term $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$where a is the first term and r is the common ratio as:
$\dfrac{S}{10}=\dfrac{{{10}^{9}}\left( {{\dfrac{11}{10}}^{10}}-1 \right)}{\dfrac{11}{10}-1}-{{11}^{10}}$
Then, by solving the above expression, we get:
\begin{align} & \dfrac{S}{10}=\dfrac{{{10}^{9}}\left( {{\dfrac{11}{10}}^{10}}-1 \right)}{\dfrac{1}{10}}-{{11}^{10}} \\ & \Rightarrow \dfrac{S}{10}={{10}^{10}} \\ & \Rightarrow S=10\times {{10}^{10}} \\ \end{align}
Now, by comparing it with the term given in the question as:
\begin{align} & 10\times {{10}^{10}}=k\times {{10}^{9}} \\ & \Rightarrow 100\times {{10}^{9}}=k\times {{10}^{9}} \\ & \Rightarrow k=100 \\ \end{align}
So, we get the value of k as 100.

So, the correct answer is “Option (c)”.

Note: Now, to solve these type of the questions we need to know some of the basic formulas for the sum of the series of the GP which the first term as a and common ratio as r with the following conditions as:
${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$ for (r>1)
${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$for( r<1)