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Question: If n is an integer greater than \(1\), then \(a - {}^n{C_1}(a - 1) + {}^n{C_2}(a - 2) - ...............

If n is an integer greater than 11, then a−nC1(a−1)+nC2(a−2)−............+(−1)n(a−n)=a - {}^n{C_1}(a - 1) + {}^n{C_2}(a - 2) - ............ + {( - 1)^n}(a - n) =
A. a B. 0 C. a2 D. 2n  A.{\text{ a}} \\\ {\text{B}}{\text{. 0}} \\\ {\text{C}}{\text{. }}{{\text{a}}^2} \\\ {\text{D}}{\text{. }}{{\text{2}}^n} \\\

Explanation

Solution

Use here expansion of the series formula and simplify using basic mathematical operations. To expand the given series use ∑r=0n(−1)rnCr(a−r)\sum\limits_{r = 0}^n {{{( - 1)}^r}{}^n{C_r}} (a - r).

Complete step by step answer:
Take the given equation -
a−nC1(a−1)+nC2(a−2)−............+(−1)n(a−n)a - {}^n{C_1}(a - 1) + {}^n{C_2}(a - 2) - ............ + {( - 1)^n}(a - n)
=∑r=0n(−1)rnCr(a−r)= \sum\limits_{r = 0}^n {{{( - 1)}^r}{}^n{C_r}} (a - r)
Now, expand the right hand side of the equation – by the factor (a-r)
=∑r=0n(−1)rnCra−∑r=0n(−1)rnCrr= \sum\limits_{r = 0}^n {{{( - 1)}^r}} {}^n{C_r}a - \sum\limits_{r = 0}^n {{{( - 1)}^r}} {}^n{C_r}r
Now put r=0r = 0
=∑r=0n(−1)rnCra−∑r=0n(−1)rnCrr‾−∑r=1n(−1)rnCrr= \sum\limits_{r = 0}^n {{{( - 1)}^r}} {}^n{C_r}a - \underline {\sum\limits_{r = 0}^n {{{( - 1)}^r}} {}^n{C_r}r} - \sum\limits_{r = 1}^n {{{( - 1)}^r}} {}^n{C_r}r
The middle term becomes zero when r=0r = 0
=∑r=0n(−1)rnCra−∑r=1n(−1)rnCrr= \sum\limits_{r = 0}^n {{{( - 1)}^r}} {}^n{C_r}a - \sum\limits_{r = 1}^n {{{( - 1)}^r}} {}^n{C_r}r
=a(1−1)n−∑r=1n(−1)rnrn−1Cr−1r= a{(1 - 1)^n} - \sum\limits_{r = 1}^n {{{( - 1)}^r}} \dfrac{n}{r}{}^{n - 1}{C_{r - 1}}r
(Using, nCr=nrn−1Cr−1{}^n{C_r} = \dfrac{n}{r}{}^{n - 1}{C_{r - 1}} )
(Also, (1+x)n=∑r=0nnCrxr{(1 + x)^n} = \sum\limits_{r = 0}^n {{}^n} {C_r}{x^r} )
As, one minus one is zero, first term becomes zero
=0−n∑r=1n(−1)rn−1Cr−1 =−n∑r=1n(−1)rn−1Cr−1  = 0 - n{\sum\limits_{r = 1}^n {( - 1)} ^r}{}^{n - 1}{C_{r - 1}} \\\ = - n{\sum\limits_{r = 1}^n {( - 1)} ^r}{}^{n - 1}{C_{r - 1}} \\\
Now, expand the above equation –
=−n[−n−1C0+n−1C1+n−1C2..........(−1)n−1n−1Cn−1]= - n[ - {}^{n - 1}{C_0} + {}^{n - 1}{C_1} + {}^{n - 1}{C_2}..........{( - 1)^{n - 1}}{}^{n - 1}{C_{n - 1}}]
As, (1−1)n−1={(1 - 1)^{n - 1}} = (1−1)n−1=n−1C0+n−1C1+n−1C2..........(−1)n−1n−1Cn−1 (1−1)n−1=0  {(1 - 1)^{n - 1}} = {}^{n - 1}{C_0} + {}^{n - 1}{C_1} + {}^{n - 1}{C_2}..........{( - 1)^{n - 1}}{}^{n - 1}{C_{n - 1}} \\\ {(1 - 1)^{n - 1}} = 0 \\\
[As one minus one is equal to zero, place the value equal to zero]
=n(0) =0  = n(0) \\\ = 0 \\\
Therefore, : If n is an integer greater than 11, then a−nC1(a−1)+nC2(a−2)−............+(−1)n(a−n)=0a - {}^n{C_1}(a - 1) + {}^n{C_2}(a - 2) - ............ + {( - 1)^n}(a - n) = 0

Hence, from the above multiple choices, the option B is the correct answer.

Note: The summation and series are the most important part of the probability theory. One should remember the basic formulas for the finite and infinite series. And remember to use applicable basic simplifications of the equations.