
If \[{}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}} - {}^n{C_r}\] , then the value of \[r\] and the minimum value of \[n\] are.
A. \[10\]
B. \[11\]
C. \[12\]
D. \[13\]
Answer
164.1k+ views
Hint: Here, the equation of the combination terms is given. First, simplify the given equation by adding \[{}^n{C_r}\] on both sides. Then, solve both sides using the combination identities. After that, equate both sides and find the quadratic equation that contains \[r\]. Solve the quadratic equation and find the value of \[r\]. In the end, use that value and find the minimum value of \[n\].
Formula Used: Combination Property: \[{}^n{C_r} = {}^{n - 1}{C_r} + {}^{n - 1}{C_{r - 1}}\]
Complete step by step solution: The given equation is \[{}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}} - {}^n{C_r}\].
Let’s simplify the equation.
Add \[{}^n{C_r}\] on both sides.
\[{}^n{C_r} + {}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}} - {}^n{C_r} + {}^n{C_r}\]
\[ \Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}}\]
Apply the identity of combination \[{}^n{C_r} = {}^{n - 1}{C_r} + {}^{n - 1}{C_{r - 1}}\] on the left-hand side.
We get,
\[{}^{2n}{C_r} + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}}\]
\[ \Rightarrow {}^{2n + 1}{C_r} = {}^{2n + 1}{C_{{r^2} - 132}}\]
Equating both sides, we get
\[r = {r^2} - 132\]
\[ \Rightarrow {r^2} - r - 132 = 0\]
\[ \Rightarrow \left( {r - 12} \right)\left( {r + 11} \right) = 0\]
\[ \Rightarrow r - 12 = 0\] or \[r + 11 = 0\]
\[ \Rightarrow r = 12\] or \[r = - 11\]
Since, the value of \[r\] cannot be negative.
Hence, \[r = 12\] is the only possible solution.
From the given equation, we get
\[n \ge r\]
\[ \Rightarrow n \ge 12\]
Thus, the minimum value of \[n\] is \[12\].
Option ‘C’ is correct
Note: The combination property \[{}^n{C_r} = {}^{n - 1}{C_r} + {}^{n - 1}{C_{r - 1}}\] is used to solve the problems that involves combination as their main concept. We can prove this combination property by solving both sides using the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
Formula Used: Combination Property: \[{}^n{C_r} = {}^{n - 1}{C_r} + {}^{n - 1}{C_{r - 1}}\]
Complete step by step solution: The given equation is \[{}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}} - {}^n{C_r}\].
Let’s simplify the equation.
Add \[{}^n{C_r}\] on both sides.
\[{}^n{C_r} + {}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}} - {}^n{C_r} + {}^n{C_r}\]
\[ \Rightarrow {}^n{C_r} + {}^n{C_{r - 1}} + {}^{n + 1}{C_{r - 1}} + {}^{n + 2}{C_{r - 1}} + .... + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}}\]
Apply the identity of combination \[{}^n{C_r} = {}^{n - 1}{C_r} + {}^{n - 1}{C_{r - 1}}\] on the left-hand side.
We get,
\[{}^{2n}{C_r} + {}^{2n}{C_{r - 1}} = {}^{2n + 1}{C_{{r^2} - 132}}\]
\[ \Rightarrow {}^{2n + 1}{C_r} = {}^{2n + 1}{C_{{r^2} - 132}}\]
Equating both sides, we get
\[r = {r^2} - 132\]
\[ \Rightarrow {r^2} - r - 132 = 0\]
\[ \Rightarrow \left( {r - 12} \right)\left( {r + 11} \right) = 0\]
\[ \Rightarrow r - 12 = 0\] or \[r + 11 = 0\]
\[ \Rightarrow r = 12\] or \[r = - 11\]
Since, the value of \[r\] cannot be negative.
Hence, \[r = 12\] is the only possible solution.
From the given equation, we get
\[n \ge r\]
\[ \Rightarrow n \ge 12\]
Thus, the minimum value of \[n\] is \[12\].
Option ‘C’ is correct
Note: The combination property \[{}^n{C_r} = {}^{n - 1}{C_r} + {}^{n - 1}{C_{r - 1}}\] is used to solve the problems that involves combination as their main concept. We can prove this combination property by solving both sides using the combination formula \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\].
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