Question

Find the value of given binomial series ${}^{n+1}{{C}_{2}}+2\left[ {}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+......+{}^{n}{{C}_{2}} \right]$.
(a) $\dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6}$
(b) $\dfrac{n\left( n+1 \right)}{2}$
(c) $\dfrac{n\left( n-1 \right)\left( 2n-1 \right)}{6}$
(d) None of these

Answer 1

Hint: In this question, we first need to look into the definitions of binomial theorem and binomial expansion. Then using the laws of binomial coefficients we can rewrite the given equation and simplify each term accordingly to get the solution.
${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$

Complete step-by-step solution -
Let us look at some of the basic definitions and terms of the binomial theorem.
Binomial Theorem for Positive Integer:
If n is any positive integer, then
${{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+......{}^{n}{{C}_{n}}{{a}^{n}}$
${{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}}{{x}^{n-r}}{{a}^{r}}$
This is called a binomial theorem.
Here,${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}}etc$are called binomial coefficients and
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$ for $0\le r\le n$
PROPERTIES OF BINOMIAL THEOREM:
Total number of terms in the expansion of (x + a)n is (n + 1).
The sum of the indices of x and a in each term is n.
The above expansion is also true when x and a are complex numbers.
The coefficient of terms equidistant from the beginning and the end are equal. These coefficients are known as the binomial coefficients and
${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$, r = 0, 1, 2, etc
The values of the binomial coefficients steadily increase to maximum and then steadily decrease.
Important result on Binomial Coefficients:
${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$
Now, let us look at the given question.
$\Rightarrow {}^{n+1}{{C}_{2}}+2\left[ {}^{2}{{C}_{2}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+......+{}^{n}{{C}_{2}} \right]$
In this we already know that
${}^{2}{{C}_{2}}=1={}^{3}{{C}_{3}}$
Now, on replacing the terms of the equation we get,
$\Rightarrow {}^{n+1}{{C}_{2}}+2\left[ {}^{3}{{C}_{3}}+{}^{3}{{C}_{2}}+{}^{4}{{C}_{2}}+......+{}^{n}{{C}_{2}} \right]$

& \Rightarrow {}^{n+1}{{C}_{2}}+2\left[ {}^{4}{{C}_{3}}+{}^{4}{{C}_{2}}+......+{}^{n}{{C}_{2}} \right] \\\ & \left[ \because {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} \right] \\\ \end{aligned}$$ Now, by applying the same formula we would finally get, $$\Rightarrow {}^{n+1}{{C}_{2}}+2\left[ {}^{n}{{C}_{3}}+{}^{n}{{C}_{2}} \right]$$ $$\begin{aligned} & \Rightarrow {}^{n+1}{{C}_{2}}+2\times {}^{n+1}{{C}_{3}} \\\ & \left[ \because {}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}} \right] \\\ \end{aligned}$$ Now, we can rewrite the above equation as follows: $$\begin{aligned} & \Rightarrow {}^{n+1}{{C}_{2}}+{}^{n+1}{{C}_{3}}+{}^{n+1}{{C}_{3}} \\\ & \Rightarrow {}^{n+2}{{C}_{3}}+{}^{n+1}{{C}_{3}} \\\ \end{aligned}$$ Let us simplify it further by using the binomial expansion formula. $$\begin{aligned} & {}^{n+2}{{C}_{3}}=\dfrac{\left( n+2 \right)!}{\left( n+2-3 \right)!3!} \\\ & {}^{n+1}{{C}_{3}}=\dfrac{\left( n+1 \right)!}{\left( n+1-3 \right)!3!} \\\ \end{aligned}$$ $$\Rightarrow \dfrac{\left( n+2 \right)!}{\left( n+2-3 \right)!3!}+\dfrac{\left( n+1 \right)!}{\left( n+1-3 \right)!3!}$$ $$\Rightarrow \dfrac{\left( n+2 \right)!}{\left( n-1 \right)!3!}+\dfrac{\left( n+1 \right)!}{\left( n-2 \right)!3!}$$ Now, by taking out the common terms we can write it as $$\Rightarrow \dfrac{\left( n+1 \right)!}{\left( n-2 \right)!3!}\left[ \dfrac{n+2}{n-1}+1 \right]$$ $$\Rightarrow \dfrac{\left( n+1 \right)\times n\times \left( n-1 \right)\times \left( n-2 \right)!}{\left( n-2 \right)!3!}\left[ \dfrac{2n+1}{n-1} \right]$$ $$\begin{aligned} & \Rightarrow \dfrac{\left( n+1 \right)\times n\times \left( 2n+1 \right)}{3!} \\\ & \Rightarrow \dfrac{n\left( n+1 \right)\left( 2n+1 \right)}{6} \\\ \end{aligned}$$ Hence, the correct option is (a). Note: Instead of using the law of binomial coefficients and simplifying the given equation we can also solve it by expanding each term but it would not be solved quickly and would take a lot of time. While simplifying the factorial terms it is preferable to take the common terms out and then solve it accordingly because we get terms common that would be cancelled and the equation converts into simplified form. $${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$$ $$\dfrac{\left( n+1 \right)!}{\left( n-2 \right)!3!}\left[ \dfrac{n+2}{n-1}+1 \right]$$