Question

Prove the given expression $\dfrac{{{C}_{1}}}{{{C}_{0}}}+2\dfrac{{{C}_{2}}}{{{C}_{1}}}+3\dfrac{{{C}_{3}}}{{{C}_{2}}}+......+n\dfrac{{{C}_{n}}}{{{C}_{n-1}}}=\dfrac{n\left( n+1 \right)}{2}$

Answer 1

Hint: Here we have to prove the given expression. First we will solve the left hand side of the given equation or expression. We will simplify each term separately by using the combination formula and then we will add the terms and after further simplifying the sum, we get the value same as the right hand side of the expression.

Complete step-by-step answer:
We know the combination formula;
${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$
The given expression is $\dfrac{{{C}_{1}}}{{{C}_{0}}}+2\dfrac{{{C}_{2}}}{{{C}_{1}}}+3\frac{{{C}_{3}}}{{{C}_{2}}}+......+n\dfrac{{{C}_{n}}}{{{C}_{n-1}}}=\dfrac{n\left( n+1 \right)}{2}$
We will consider the left hand side of the expression first.
Therefore, we have
$=\dfrac{{{C}_{1}}}{{{C}_{0}}}+2\dfrac{{{C}_{2}}}{{{C}_{1}}}+3\frac{{{C}_{3}}}{{{C}_{2}}}+......+n\dfrac{{{C}_{n}}}{{{C}_{n-1}}}$
Here ${{C}_{1}}$ means ${}^{n}{{C}_{1}}$, similarly the other terms are like that only.
Applying the combination formula for each term, we get
$=\dfrac{\dfrac{n!}{1!\times \left( n-1 \right)!}}{\dfrac{n!}{0!\times n!}}+2\dfrac{\dfrac{n!}{2!\times \left( n-2 \right)!}}{\dfrac{n!}{1!\times \left( n-1 \right)!}}+3\dfrac{\dfrac{n!}{3!\times \left( n-3 \right)!}}{\dfrac{n!}{2!\times \left( n-2 \right)!}}+......+n\dfrac{\dfrac{n!}{n!\times \left( n-n \right)!}}{\dfrac{n!}{\left( n-1 \right)!\times \left( n-\left( n-1 \right) \right)!}}$
On simplifying the terms, we get
$=\dfrac{n!}{1\times \left( n-1 \right)!}+2\dfrac{\left( n-1 \right)!}{2!\times \left( n-2 \right)!}+3\dfrac{2!\times \left( n-2 \right)!}{3!\times \left( n-3 \right)!}+......+n\dfrac{\left( n-1 \right)!}{1!\times n!}$
On finding the factorials of each term, we get
$=\dfrac{n}{1}+2\dfrac{n-1}{2\times 1}+3\dfrac{2\times 1\times \left( n-2 \right)}{3\times 2\times 1}+......+n\times \dfrac{1}{n}$
Multiplying the terms in numerator and denominator of each term, we get
$=n+\left( n-1 \right)+\left( n-2 \right)+......+1$ ………. $\left( 1 \right)$
This series is forming an AP because the difference of each term from the preceding term is 1 which is constant. We will apply the formula of sum of an AP.
We know the formula of sum of an AP is given by
$\Rightarrow {{S}_{n}}=n\times \dfrac{a+l}{2}$
Here $n$ is the number of terms in an AP, $a$ is the first term of an AP and $l$ is the last term of an AP.
We will apply the formula of sum of an AP in equation 1.
Here first term is $n$, number of terms in the AP is $n$ and the last term is
Substituting these values in the formula of sum of an AP, we get
$=n\times \dfrac{\left( n+1 \right)}{2}=\dfrac{n\left( n+1 \right)}{2}$
This value is equal to the right hand side of the given equation.
Hence, we have proved the given expression.

Note: We have obtained the sum of an arithmetic progression. An arithmetic progression is defined as a sequence in which the difference between the term and the preceding term is constant or in other words, we can say that an arithmetic progression is a sequence such that every element after the first is obtained by adding a constant term to the preceding element.