Question

The sum of the binomial of the coefficient of the 3 rd , 4 th term from the beginning and from the end of ${{(a+x)}^{n}}$ is $440$ then $n$ is
A. $10$
B. $11$
C. $12$
D. $13$

Answer 1

According to the question, to find the sum of the binomial of the coefficient we use the formula of sum as:
$C_{1}^{n}+C_{2}^{n}+C_{3}^{n}+..+C_{n-2}^{n}+C_{n-3}^{n}+..+C_{n}^{n}$
Now we only need the second and third term on both the front and back side of the above sum series and equate it with $440$. After that change those terms in factorial form and frame a cubic equation with $440$.

Complete step by step solution:
As given in the question, the second, third terms from the beginning and from the end are used. Now the sum of the series is given as:
$\Rightarrow C_{1}^{n}+C_{2}^{n}+C_{3}^{n}+..+C_{n-2}^{n}+C_{n-3}^{n}+..+C_{n}^{n}$
And the second and the third term from the sum of the series from the beginning is given as:
$\Rightarrow C_{2}^{n}+C_{3}^{n}$
And the sum of the second and the third term from the last is given as:
$\Rightarrow C_{n-2}^{n}+C_{n-3}^{n}$
Now converting the terms in form of factorial $n$, we get the value of the sum of the combination for the second and third from both first and the last term is:
$\Rightarrow C_{2}^{n}+C_{3}^{n}+C_{n-2}^{n}+C_{n-3}^{n}=440$
Now when writing terms in binomial we can say that the values at the start and the end are one and the same thing meaning let us write the factorial form of $C_{2}^{n}$ we get the factorial form as:
$C_{2}^{n}=\dfrac{n!}{2!\left( n-2 \right)!}$
And if we find the factorial term of $C_{n-2}^{n}$ we get the form as:
$C_{n-2}^{n}=\dfrac{n!}{\left( n-2 \right)!\left( 2 \right)!}$
Therefore, we can see that both the values of $C_{2}^{n},C_{n-2}^{n}$ are the same.
Hence, the equation of the sum
$\Rightarrow C_{2}^{n}+C_{3}^{n}+C_{n-2}^{n}+C_{n-3}^{n}=440$ can be rewritten as:
$\Rightarrow C_{2}^{n}+C_{3}^{n}+C_{2}^{n}+C_{3}^{n}=440$
Hence, placing the factorial form in the above equation we get the value as:
$\Rightarrow C_{2}^{n}+C_{3}^{n}+C_{2}^{n}+C_{3}^{n}=440$
$\Rightarrow 2\left( \dfrac{n!}{2!\left( n-2 \right)!}+\dfrac{n!}{3!\left( n-3 \right)!} \right)=440$
Solving the above equation so as to find a cubic equation from which we can get the value of $n$ as:
$\Rightarrow \dfrac{n\left( n-1 \right)}{2}\dfrac{\left( n+1 \right)}{3}=220$
$\Rightarrow \dfrac{\left( {{n}^{2}}-1 \right)n}{6}=220$
Simplifying the above equation we get the value of the simplified equation as:
$\Rightarrow {{n}^{3}}-n=1320$
$\Rightarrow {{n}^{3}}+11{{n}^{2}}+120n-11{{n}^{2}}-121n-1320$
$\Rightarrow n\left( {{n}^{2}}+11n+120 \right)-11\left( {{n}^{2}}-11n-120 \right)$
Now grouping the factors, we get the simplified value as:
$\Rightarrow \left( n-11 \right)\left( {{n}^{2}}+11n+120 \right)$
Now equating the one of the value of the above equation equal to zero we get the value as:
$\Rightarrow \left( n-11 \right)=0$
$\Rightarrow n=11$
Therefore, the value of $n$ is given as $11$.

Note: The value of $C_{2}^{n}$ and $C_{n-2}^{n}$ are same but how let us expand the form of $C_{n-2}^{n}$ into factorial form:
Now the formula of the factorial terms is given as:
$C_{r}^{n}=\dfrac{n!}{\left( r \right)!\left( n-r \right)!}$
Now placing the value of $r$ as $n-2$, we get the term as:
$C_{n-2}^{n}=\dfrac{n!}{\left( n-2 \right)!\left( n-\left( n-2 \right) \right)!}$
Now if we remove the bracket of $n-\left( n-2 \right)$ we get $2$ and placing it in the above term we get the value as:
$C_{n-2}^{n}=\dfrac{n!}{\left( n-2 \right)!\left( 2 \right)!}$