Question

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

Answer 1

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.