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Question: Find the value of k for the following question. \[{{\sum\limits_{i=1}^{20}{\left( \dfrac{^{20}{{C}...

Find the value of k for the following question.
∑i=120(20Ci−120Ci+ 20Ci−1)3=k21{{\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

Explanation

Solution

First, we use nCr+ nCr−1= n+1Cr^{n}{{C}_{r}}+{{\text{ }}^{n}}{{C}_{r-1}}={{\text{ }}^{n+1}}{{C}_{r}} to simplify 20Ci+ 20Ci−1^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}} to give us 21Ci.^{21}{{C}_{i}}. Then we will expand 20Ci−121Ci\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}} using the combination formula nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} to get the simplified value. At last, we will use ∑n3=[n(n+1)2]2{{\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 k21\dfrac{k}{21} to get the value of k.

Complete step-by-step solution:
We are given that ∑i=120(20Ci−120Ci+ 20Ci−1)3=k21,{{\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 nCr^{n}{{C}_{r}} and nCr−1^{n}{{C}_{r-1}} is given as n+1Cr.^{n+1}{{C}_{r}}. That means,
nCr+ nCr−1= n+1Cr^{n}{{C}_{r}}+{{\text{ }}^{n}}{{C}_{r-1}}={{\text{ }}^{n+1}}{{C}_{r}}
So, using this, we get,
20Ci+ 20Ci−1= 20+1Ci^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}={{\text{ }}^{20+1}}{{C}_{i}}
⇒20Ci+ 20Ci−1= 21Ci{{\Rightarrow }^{20}}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}={{\text{ }}^{21}}{{C}_{i}}
Now, we have got, 20Ci−120Ci+ 20Ci−1\dfrac{^{20}{{C}_{i-1}}}{^{20}{{C}_{i}}+{{\text{ }}^{20}}{{C}_{i-1}}} as 20Ci−121Ci.\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}. Now, we will simplify 20Ci−121Ci.\dfrac{^{20}{{C}_{i-1}}}{^{21}{{C}_{i}}}. We know that nCr^{n}{{C}_{r}} is given as n!r!(n−r)!,\dfrac{n!}{r!\left( n-r \right)!}, so,
20Ci−121Ci=20!(i−1)!(20−i+1)!21!i!(21−i)!\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,
⇒20Ci−121Ci=20!×i!×(21−i)!21!×(i−1)!×(21−i)!\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,
⇒121×i\Rightarrow \dfrac{1}{21}\times i
⇒i21\Rightarrow \dfrac{i}{21}
Now, we get
⇒∑i=120(20Ci−120Ci+ 20Ci−1)3=∑i=120(i21)3[From (i)]\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 (21)3{{\left( 21 \right)}^{3}} out, we get,
=1(21)3∑i=2120(i)3=\dfrac{1}{{{\left( 21 \right)}^{3}}}\sum\limits_{i=21}^{20}{{{\left( i \right)}^{3}}}
We know that, ∑i=1n(n)3=[n(n+1)2]2\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,
=1(21)3[20×(20+1)2]2=\dfrac{1}{{{\left( 21 \right)}^{3}}}{{\left[ \dfrac{20\times \left( 20+1 \right)}{2} \right]}^{2}}
Simplifying further, we get,
=10021=\dfrac{100}{21}
So comparing 10021\dfrac{100}{21} with k21\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,
∑i=120(i21)3=1(21)3∑i=120(i)3\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 nCr^{n}{{C}_{r}} is n!r!(n−r)!.\dfrac{n!}{r!\left( n-r \right)!}. So, do not make mistake by taking nCr^{n}{{C}_{r}} as n!(n−r)!\dfrac{n!}{\left( n-r \right)!} as it gives is the wrong solution.