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Question: If n is an integer greater than 1, then \(a - {}^n{c_1}\left( {a - 1} \right) + {}^n{c_2}\left( {a -...

If n is an integer greater than 1, then a−nc1(a−1)+nc2(a−2)−.......+(−1)n(a−n)=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) a2{a^2}
D) 2n{2^n}

Explanation

Solution

It is given in the question that n is an integer greater than 1, then a−nc1(a−1)+nc2(a−2)−.......+(−1)n(a−n)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 ∑r=0n(−1)rnCr(a−r)\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−nc1(a−1)+nc2(a−2)−.......+(−1)n(a−n)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−nc1(a−1)+nc2(a−2)−.......+(−1)n(a−n)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
=∑r=0n(−1)rnCr(a−r)= \sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}{}^n{C_r}} \left( {a - r} \right)
=∑r=0n(−1)rnCr.a−∑r=0n(−1)rnCr.r= \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∑r=0n(−1)rnCr−∑r=1n(−1)rnCr.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=0r = 0 , we get 0)
Now, using formula (nCr=nrn−1Cr−1)(1+x)n=∑r=0nnCrxr\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=−1x = - 1 ⇒(1−1)n=∑r=0nnCr(−1)r \Rightarrow {\left( {1 - 1} \right)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}} {\left( { - 1} \right)^r}
=a(1−1)n−∑r=1n(−1)r.nrn−1Cr−1.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(0)−∑r=1n(−1)rn.n−1Cr−1= a\left( 0 \right) - \sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^r}n.{}^{n - 1}{C_{r - 1}}}
=0−n∑r=1n(−1)rn−1Cr−1= 0 - n\sum\limits_{r = 1}^n {{{\left( { - 1} \right)}^r}{}^{n - 1}{C_{r - 1}}}
=−n[−n−1Co+n−1C1−n−1C2.............(1)n−1n−1Cn−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(0) =0= - n\left( 0 \right) \\\ =0

Therefore, if n is an integer greater than 1, then a−nc1(a−1)+nc2(a−2)−.......+(−1)n(a−n)=0a - {}^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.
nCr=n!r!(n−r)!{}^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.
nPr=n!(n−r)!{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}