Question

If ${{\left( 20 \right)}^{19}}+2\left( 21 \right){{\left( 20 \right)}^{18}}+3{{\left( 21 \right)}^{2}}{{\left( 20 \right)}^{17}}+.......+20{{\left( 21 \right)}^{19}}=k{{\left( 20 \right)}^{19}}$ then $k$ is equal to:
(a) 400
(b) 100
(c) 441
(d) 420

Answer 1

Here in this question we have been asked to find the value of $k$ in the given equation ${{\left( 20 \right)}^{19}}+2\left( 21 \right){{\left( 20 \right)}^{18}}+3{{\left( 21 \right)}^{2}}{{\left( 20 \right)}^{17}}+.......+20{{\left( 21 \right)}^{19}}=k{{\left( 20 \right)}^{19}}$ . To answer this question, we will take out ${{\left( 20 \right)}^{19}}$ in common from the left side of the expression and then simplify the remaining part of the expression.

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of $k$ in the given equation ${{\left( 20 \right)}^{19}}+2\left( 21 \right){{\left( 20 \right)}^{18}}+3{{\left( 21 \right)}^{2}}{{\left( 20 \right)}^{17}}+.......+20{{\left( 21 \right)}^{19}}=k{{\left( 20 \right)}^{19}}$ .
Now we will take out ${{\left( 20 \right)}^{19}}$ as common from the left side of the expression to simplify the expression.
By doing that we will have $\Rightarrow {{\left( 20 \right)}^{19}}\left[ 1+2\left( \dfrac{21}{20} \right)+3{{\left( \dfrac{21}{20} \right)}^{2}}+......+20{{\left( \dfrac{21}{20} \right)}^{19}} \right]=k{{\left( 20 \right)}^{19}}$ .
Hence we can say that
$\Rightarrow 1+2\left( \dfrac{21}{20} \right)+3{{\left( \dfrac{21}{20} \right)}^{2}}+......+20{{\left( \dfrac{21}{20} \right)}^{19}}=k$ ………………………….(1)
Now we will multiply this expression with $\left( \dfrac{21}{20} \right)$ after doing that we will have $\Rightarrow k\left( \dfrac{21}{20} \right)=\left( \dfrac{21}{20} \right)+2{{\left( \dfrac{21}{20} \right)}^{2}}+3{{\left( \dfrac{21}{20} \right)}^{3}}+......+20{{\left( \dfrac{21}{20} \right)}^{20}}$ ………………….(2)
Now we will simplify the expression by subtracting eq.2 from eq.1 as follows:
$\Rightarrow k\left( 1-\dfrac{21}{20} \right)=1+\left( \dfrac{21}{20} \right)+{{\left( \dfrac{21}{20} \right)}^{2}}+......-20{{\left( \dfrac{21}{20} \right)}^{20}}$ .
Now we can say that the following part of the expression $1+\left( \dfrac{21}{20} \right)+{{\left( \dfrac{21}{20} \right)}^{2}}+......+{{\left( \dfrac{21}{20} \right)}^{19}}$ is in geometric progression.
From the basic concepts of progression we know that the sum of the $n$ terms in a geometric progression will be given as $\dfrac{a{{r}^{n}}-1}{r-1}$ where $a$ is the first term and $r$ is the $ratio$ .
Now we will have
$\Rightarrow 1+\left( \dfrac{21}{20} \right)+{{\left( \dfrac{21}{20} \right)}^{2}}+......+{{\left( \dfrac{21}{20} \right)}^{19}}=\dfrac{{{\left( \dfrac{21}{20} \right)}^{20}}-1}{\left( \dfrac{21}{20} \right)-1}$ .
Now by simplifying this we will get
$\Rightarrow \dfrac{{{\left( \dfrac{21}{20} \right)}^{20}}-1}{\dfrac{1}{20}}=20\left( {{\left( \dfrac{21}{20} \right)}^{20}}-1 \right)$ .
Now we can say that
$\begin{aligned} & \Rightarrow k\left( 1-\dfrac{21}{20} \right)=20\left( {{\left( \dfrac{21}{20} \right)}^{20}}-1 \right)-20{{\left( \dfrac{21}{20} \right)}^{20}} \\\ & \Rightarrow \dfrac{-k}{20}=-20 \\\ \end{aligned}$ .
Hence we can conclude that the value of $k$ is 400.
Therefore we will mark the option “a” as correct.

Note: While answering questions of this type we should be very careful with each and every step we are performing. Very few mistakes are possible in questions of this type and they are generally related to calculations.