Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# 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
Total views: 349.8k
Views today: 5.49k
Verified
349.8k+ views
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
${}^{n}{{C}_{n}}=1$.
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
${}^{n}{{C}_{r}}+{}^{n}{{C}_{r+1}}={}^{n+1}{{C}_{r+1}}$.
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
${}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}={}^{5}{{C}_{3}}+{}^{5}{{C}_{2}}+...+{}^{n}{{C}_{2}}$
We can reduce the above equation in the same way, to get
${}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}={}^{n}{{C}_{3}}+{}^{n}{{C}_{2}}$
And hence, we get
${}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+...+{}^{n}{{C}_{2}}={}^{n+1}{{C}_{3}}$.
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
\begin{align} & k=1 \\ & m=1 \\ & p=2 \\ & h=6 \\ \end{align}
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.