Question

If n is an integer greater than 1, then $a - {}^n{c_1}\left( {a - 1} \right) + {}^n{c_2}\left( {a - 2} \right) - ....... + {\left( { - 1} \right)^n}\left( {a - n} \right) = $
A) a
B) 0
C) ${a^2}$
D) ${2^n}$

Answer 1

Hint:
It is given in the question that n is an integer greater than 1, then $a - {}^n{c_1}\left( {a - 1} \right) + {}^n{c_2}\left( {a - 2} \right) - ....... + {\left( { - 1} \right)^n}\left( {a - n} \right)$
Firstly, expand the given series using the formula \[\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} \left( {a - r} \right)\] .
Then, solve the series further to get the required answer.

Complete step by step solution:
It is given in the question that n is an integer greater than 1, then $a - {}^n{c_1}\left( {a - 1} \right) + {}^n{c_2}\left( {a - 2} \right) - ....... + {\left( { - 1} \right)^n}\left( {a - n} \right)$
First, take the given equation:
 $a - {}^n{c_1}\left( {a - 1} \right) + {}^n{c_2}\left( {a - 2} \right) - ....... + {\left( { - 1} \right)^n}\left( {a - n} \right)$
Now, we can write above sequence as
 \[ = \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} \left( {a - r} \right)\]
 \[ = \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} .a - \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} .r\]
 \[ = a\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} - \sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} .r\] (where \[r = 0\] , we get 0)
Now, using formula \[\left( {{}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} \right){\left( {1 + x} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {x^r}\] when \[x = - 1\] \[ \Rightarrow {\left( {1 - 1} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {\left( { - 1} \right)^r}\]
 \[ = a{\left( {1 - 1} \right)^n} - \sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^r}.\dfrac{n}{r}{}^{n - 1}{C_{r - 1}}} .r\]
 \[ = a\left( 0 \right) - \sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^r}n.{}^{n - 1}{C_{r - 1}}} \]
 \[ = 0 - n\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^r}{}^{n - 1}{C_{r - 1}}} \]
 \[ = - n\left[ { - {}^{n - 1}{C_o} + {}^{n - 1}{C_1} - {}^{n - 1}{C_2}.............{{\left( 1 \right)}^{n - 1}}{}^{n - 1}{C_{n - 1}}} \right]\]
 $ = - n\left( 0 \right) \\
=0$

Therefore, if n is an integer greater than 1, then $a - {}^n{c_1}\left( {a - 1} \right) + {}^n{c_2}\left( {a - 2} \right) - ....... + {\left( { - 1} \right)^n}\left( {a - n} \right) = 0$.

Note:
Combination: A combination is a selection of items from a collection, such that the order of selection does not matter.
 ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
Permutation: permutation means arranging all the members of a set into some sequence or order.
 ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$