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The value of the expression ${}^{n+1}{{C}_{2}}+2\left[ {}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}} \right]$ is $\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$. Find $k+m+p+h$.

Last updated date: 24th Jul 2024
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Hint: First, we must reduce the expression ${}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}$ by using the property ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r+1}}$ and the fact that ${}^{n}{{C}_{n}}=1$. Then, we can evaluate the given equation using the definition ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. By comparing, we can find the value of k, m, p and h.

Complete step-by-step solution:
We all know very well that the combination of r distinct objects from a set of n distinct objects is defined as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
For r = n, we can write that ${}^{n}{{C}_{n}}=\dfrac{n!}{n!\left( n-n \right)!}$.
Hence, we get ${}^{n}{{C}_{n}}=\dfrac{1}{0!}$.
We know that the factorial of 0 is defined as 1. Thus, we can write that
Thus, we now have ${}^{2}{{C}_{2}}={}^{3}{{C}_{3}}=1$.
Hence, we can write
${}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}={}^{3}{{C}_{3}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}...\left( i \right)$
We know the property that
Hence, we can say that ${}^{3}{{C}_{2}}+{}^{3}{{C}_{3}}={}^{4}{{C}_{3}}$.
Thus, using the above value on the right hand side of equation (i), we get
${}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}={}^{4}{{C}_{3}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}...\left( ii \right)$
Using the same property again, we can write ${}^{4}{{C}_{2}}+{}^{4}{{C}_{3}}={}^{5}{{C}_{3}}$.
Thus, using the above value on the right hand side of equation (ii), we get
We can reduce the above equation in the same way, to get
And hence, we get
So, we can write the given equation as
${}^{n+1}{{C}_{2}}+2\left[ {}^{n+1}{{C}_{3}} \right]=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$
We can write the above equation as
${}^{n+1}{{C}_{2}}+{}^{n+1}{{C}_{3}}+{}^{n+1}{{C}_{3}}=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$
Using the property ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r+1}}$ on the left hand side of above equation, we can write
${}^{n+2}{{C}_{3}}+{}^{n+1}{{C}_{3}}=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$.
Using the definition of combinations, we can write
$\dfrac{\left( n+2 \right)!}{3!\left( n+2-3 \right)!}+\dfrac{\left( n+1 \right)!}{3!\left( n+1-3 \right)!}=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$.
On simplification, we get
$\dfrac{\left( n+2 \right)!}{3!\left( n-1 \right)!}+\dfrac{\left( n+1 \right)!}{3!\left( n-2 \right)!}=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$.
Hence, we can now write
$\dfrac{\left( n+2 \right)\left( n+1 \right)n}{6}+\dfrac{\left( n+1 \right)n\left( n-1 \right)}{6}=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$.
We can simplify this equation as
$\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}=\dfrac{n\left( n+k \right)\left( pn+m \right)}{h}$
On comparing, we can write
  & k=1 \\
 & m=1 \\
 & p=2 \\
 & h=6 \\
Thus, $k+m+p+h=1+1+2+6$.
Hence, the value of $k+m+p+h$ is 10.

Note: We must remember the property of combination, ${}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r+1}}$ by heart, as it is very easy to make a mistake in writing this property. Also, we must not try to simplify the given equation by solving each combination separately, as this will make the equation much more complex.