Question

Solve \[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + ....... + n\dfrac{{{C_n}}}{{{C_{n - 1}}}} = \dfrac{{n\left( {n + 1} \right)}}{2}\].

Answer 1

Hint: In the given questions we have to prove that the LHS term can be simplified as the term on RHS , so we will solve LHS . In this we will use the formula for combination which is \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] . Also , we can write \[{C_r}\] as \[^n{C_r}\] . Since it is a series therefore , we will use \[\sum {} \] ( summation ) for simplification.

Complete step by step answer:
Given : \[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + ....... + n\dfrac{{{C_n}}}{{{C_{n - 1}}}}\]
Now the above series can be simplified as :
\[ \sum\limits_{r = 1}^n {r \times \dfrac{{^n{C_r}}}{{^n{C_{r - 1}}}}} .....eqn(1)\]
Here , \[r\] represents the natural numbers which are multiplied with each term up to \[n\] . The \[^n{C_r}\] represents the numerator of the every term up to \[{C_n}\] and the \[^n{C_{r - 1}}\] represents the denominator of each term up to \[{C_{n - 1}}\] .
Now using the formula \[^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] in eqn (1) we get ,
\[\sum\limits_{r = 1}^n {r \times \dfrac{{\dfrac{{n!}}{{\left( {n - r} \right)!r!}}}}{{\dfrac{{n!}}{{\left( {n - \left( {r - 1} \right)} \right)!\left( {r - 1} \right)!}}}}} \]

On simplifying we get ,
\[\sum\limits_{r = 1}^n {r \times \dfrac{{\dfrac{{n!}}{{\left( {n - r} \right)!r!}}}}{{\dfrac{{n!}}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}}}} \]
On cancelling out the term \[n!\] we get ,
\[\sum\limits_{r = 1}^n {r \times \dfrac{{\dfrac{1}{{\left( {n - r} \right)!r!}}}}{{\dfrac{1}{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}}}} \]
On simplifying we get ,
\[\sum\limits_{r = 1}^n {r \times \dfrac{{\left( {n - r + 1} \right)!\left( {r - 1} \right)!}}{{\left( {n - r} \right)!r!}}} \]
Now expanding the factorials we get ,
\[ \sum\limits_{r = 1}^n {r \times \dfrac{{\left( {n - r + 1} \right) \times \left( {n - r} \right)!\left( {r - 1} \right)!}}{{\left( {n - r} \right)! \times r \times \left( {r - 1} \right)!}}} \]

Now cancelling out the terms from numerator and denominator we get ,
\[\sum\limits_{r = 1}^n {\left( {n - r + 1} \right)} \]
On simplifying we get ,
\[\sum\limits_{r = 1}^n {\left( {n + 1} \right)} - \sum\limits_{r = 1}^n r \]
On solving we get ,
\[\left( {n + 1} \right)\sum\limits_{r = 1}^n 1 - \sum\limits_{r = 1}^n r \]
Now , in above expression the \[1\] is up to \[n\] times so it can be written as \[n\] and the second term is the sum of natural number which is equals to \[\dfrac{{n\left( {n + 1} \right)}}{2}\] . So applying these we get ,
\[\left[ {\left( {n + 1} \right) \times n} \right] - \left( {1 + 2 + 3 + 4 + ...... + n} \right)\]
On simplifying we get
\[n\left( {n + 1} \right) - \dfrac{{n\left( {n + 1} \right)}}{2}\]
On solving we get ,
 \[\therefore \dfrac{{n\left( {n + 1} \right)}}{2}\]
Hence proved.

Note:In these types of questions try to solve them or simplify them using summation as you do not have to solve the series for every term as it is a complex way instead simplify them to general form or term . Moreover remember the formula for combination and factorial of a number.