Question

What is the value of the expression given below?
\[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}}\]
(a) $\dfrac{{n(n + 1)}}{2}$
(b) \[\dfrac{{{n^2}}}{4}\]
(c) $\dfrac{2}{{n(n + 1)}}$
(d) $n + 1$

Answer 1

Hint: We are going to find the desired value for this problem by using the basic concept of permutation and combinations, factorials, etc. and then simplifying it to a number of iterations that is up to the nth terms of the equations.

Complete step-by-step answer:
$\because $The given expression for the combinations of the equations can be written as,\[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots \ldots \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}}\]
Predominantly the given equation can also be extracted as (in a generalised manner), the equation becomes
\[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots \ldots \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}} = \dfrac{n}{1} + 2\dfrac{{n(n - 1)}}{{2!}} \times \dfrac{1}{n} + 3\dfrac{{n(n - 1)(n - 2)}}{{3!}} \times \dfrac{1}{{\dfrac{{n(n - 1)}}{{2!}}}} + \ldots \ldots \ldots + \dfrac{{n \times 1}}{n}\]As a result, solving the equation mathematically, we get
\[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots \ldots \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}} = n + (n - 1) + (n - 2) + \ldots \ldots \ldots + 1\]
As a result, the infinite equation can be written as
\[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots \ldots \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}} = 1 + 2 + 3 + \ldots \ldots \ldots + n\]
Now, hence the generalised solution of the equation is formed as
\[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots \ldots \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}} = \dfrac{{n(n + 1)}}{2}\]
Therefore, the required answer is $\dfrac{{n(n + 1)}}{2}$respectively.
$\therefore \Rightarrow $The correct option is (a)
So, the correct answer is “Option a”.

Note: While solving the solution one must remember the fundamental terminology regarding the algebraic (general) terms where they have been used here in the respective solution. For an instance, the existing solution depends on the formula \[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + \ldots \ldots \ldots + n\dfrac{{{C_n}}}{{{C_{n - 1}}}} = \dfrac{n}{1} + 2\dfrac{{n(n - 1)}}{{2!}} \times \dfrac{1}{n} + 3\dfrac{{n(n - 1)(n - 2)}}{{3!}} \times \dfrac{1}{{\dfrac{{n(n - 1)}}{{2!}}}} + \ldots \ldots \ldots + \dfrac{{n \times 1}}{n}\] itself, which in term are one of the generalised formulae of combinational statement or equation. Clarifying the fundamental formulae one must remember the rules of permutation and combination concept (such as $0! = 1$,\[{}^n{c_n} = 1\],\[{}^n{c_1} = n\]) which is useful for the ease of the problem.