Solveeit Logo
Login

Question

Question: Solve \[\dfrac{{{C_1}}}{{{C_0}}} + 2\dfrac{{{C_2}}}{{{C_1}}} + 3\dfrac{{{C_3}}}{{{C_2}}} + ....... +...

Solve C1C0+2C2C1+3C3C2+.......+nCnCn−1=n(n+1)2\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}.

Explanation

Solution

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 nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} . Also , we can write Cr{C_r} as nCr^n{C_r} . Since it is a series therefore , we will use ∑\sum {} ( summation ) for simplification.

Complete step by step answer:
Given : C1C0+2C2C1+3C3C2+.......+nCnCn−1\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 :
∑r=1nr×nCrnCr−1.....eqn(1)\sum\limits_{r = 1}^n {r \times \dfrac{{^n{C_r}}}{{^n{C_{r - 1}}}}} .....eqn(1)
Here , rr represents the natural numbers which are multiplied with each term up to nn . The nCr^n{C_r} represents the numerator of the every term up to Cn{C_n} and the nCr−1^n{C_{r - 1}} represents the denominator of each term up to Cn−1{C_{n - 1}} .
Now using the formula nCr=n!(n−r)!r!^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}} in eqn (1) we get ,
∑r=1nr×n!(n−r)!r!n!(n−(r−1))!(r−1)!\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 ,
∑r=1nr×n!(n−r)!r!n!(n−r+1)!(r−1)!\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!n! we get ,
∑r=1nr×1(n−r)!r!1(n−r+1)!(r−1)!\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 ,
∑r=1nr×(n−r+1)!(r−1)!(n−r)!r!\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 ,
∑r=1nr×(n−r+1)×(n−r)!(r−1)!(n−r)!×r×(r−1)!\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 ,
∑r=1n(n−r+1)\sum\limits_{r = 1}^n {\left( {n - r + 1} \right)}
On simplifying we get ,
∑r=1n(n+1)−∑r=1nr\sum\limits_{r = 1}^n {\left( {n + 1} \right)} - \sum\limits_{r = 1}^n r
On solving we get ,
(n+1)∑r=1n1−∑r=1nr\left( {n + 1} \right)\sum\limits_{r = 1}^n 1 - \sum\limits_{r = 1}^n r
Now , in above expression the 11 is up to nn times so it can be written as nn and the second term is the sum of natural number which is equals to n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2} . So applying these we get ,
[(n+1)×n]−(1+2+3+4+......+n)\left[ {\left( {n + 1} \right) \times n} \right] - \left( {1 + 2 + 3 + 4 + ...... + n} \right)
On simplifying we get
n(n+1)−n(n+1)2n\left( {n + 1} \right) - \dfrac{{n\left( {n + 1} \right)}}{2}
On solving we get ,
∴n(n+1)2\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.