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# Find the value of k for the following question.${{\sum\limits_{i=1}^{20}{\left( \dfrac{^{20}{{C}_{i-1}}}{^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}} \right)}}^{3}}=\dfrac{k}{21}$(a) 200(b) 50(c) 100(d) 400

Last updated date: 12th Sep 2024
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Hint: First, we use $^{n}{{C}_{r}}+{{\text{ }}^{n}}{{C}_{r-1}}={{\text{ }}^{n+1}}{{C}_{r}}$ to simplify $^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}$ to give us $^{21}{{C}_{i}}.$ Then we will expand $\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}$ using the combination formula $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ to get the simplified value. At last, we will use ${{\sum{{{n}^{3}}=\left[ \dfrac{n\left( n+1 \right)}{2} \right]}}^{2}}$ to get our final answer. Once we have our solution, we will compare it with $\dfrac{k}{21}$ to get the value of k.

Complete step-by-step solution:
We are given that ${{\sum\limits_{i=1}^{20}{\left( \dfrac{^{20}{{C}_{i-1}}}{^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}} \right)}}^{3}}=\dfrac{k}{21},$ we will find the value of k. We know a rule of combination which says the sum of $^{n}{{C}_{r}}$ and $^{n}{{C}_{r-1}}$ is given as $^{n+1}{{C}_{r}}.$ That means,
$^{n}{{C}_{r}}+{{\text{ }}^{n}}{{C}_{r-1}}={{\text{ }}^{n+1}}{{C}_{r}}$
So, using this, we get,
$^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}={{\text{ }}^{20+1}}{{C}_{i}}$
${{\Rightarrow }^{20}}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}={{\text{ }}^{21}}{{C}_{i}}$
Now, we have got, $\dfrac{^{20}{{C}_{i-1}}}{^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}}$ as $\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}.$ Now, we will simplify $\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}.$ We know that $^{n}{{C}_{r}}$ is given as $\dfrac{n!}{r!\left( n-r \right)!},$ so,
$\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}=\dfrac{\dfrac{20!}{\left( i-1 \right)!\left( 20-i+1 \right)!}}{\dfrac{21!}{i!\left( 21-i \right)!}}$
Simplifying, we get,
$\Rightarrow \dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}=\dfrac{20!\times i!\times \left( 21-i \right)!}{21!\times \left( i-1 \right)!\times \left( 21-i \right)!}$
Cancelling the like terms, we get,
$\Rightarrow \dfrac{1}{21}\times i$
$\Rightarrow \dfrac{i}{21}$
Now, we get
$\Rightarrow {{\sum\limits_{i=1}^{20}{\left( \dfrac{^{20}{{C}_{i-1}}}{^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}} \right)}}^{3}}=\sum\limits_{i=1}^{20}{{{\left( \dfrac{i}{21} \right)}^{3}}\left[ \text{From (i)} \right]}$
Takin ${{\left( 21 \right)}^{3}}$ out, we get,
$=\dfrac{1}{{{\left( 21 \right)}^{3}}}\sum\limits_{i=21}^{20}{{{\left( i \right)}^{3}}}$
We know that, $\sum\limits_{i=1}^{n}{{{\left( n \right)}^{3}}}={{\left[ \dfrac{n\left( n+1 \right)}{2} \right]}^{2}}$ so as n = 20, we get,
$=\dfrac{1}{{{\left( 21 \right)}^{3}}}{{\left[ \dfrac{20\times \left( 20+1 \right)}{2} \right]}^{2}}$
Simplifying further, we get,
$=\dfrac{100}{21}$
So comparing $\dfrac{100}{21}$ with $\dfrac{k}{21}$ we get k = 100.
Hence, the option (c) is the right answer.

Note: Remember that subtraction given to us is over i. So, we can easily take out the terms which are free from i. So, using this fact, we get,
$\sum\limits_{i=1}^{20}{{{\left( \dfrac{i}{21} \right)}^{3}}=\dfrac{1}{{{\left( 21 \right)}^{3}}}\sum\limits_{i=1}^{20}{{{\left( i \right)}^{3}}}}$
The formula of $^{n}{{C}_{r}}$ is $\dfrac{n!}{r!\left( n-r \right)!}.$ So, do not make mistake by taking $^{n}{{C}_{r}}$ as $\dfrac{n!}{\left( n-r \right)!}$ as it gives is the wrong solution.