Question

If $(1 + x) = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2}...... + {}^n{C_n}{x^n}\,,\,x \in N$ then prove that ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^{m - 1}} \times {}^n{C_{m - 1}}\,$ = ${( - 1)^{m - 1}}\dfrac{{(n - 1)(n - 2)........(n - m + 1)}}{{(m - 1)!}}$.

Answer 1

We can solve this question by using induction method. Which means we can test first taking m=1 and then m=2 then we will assume that it will be satisfied till m=a and then we will prove that it will also be satisfied for m=a+1. For this we need to know that if we place x as -1 in the given expansion of the question we can get the equation for what we were asked to prove.

Complete step by step answer:
Now let us take m=1
Which implies the LHS will be ${}^n{C_0}$which is 1 and the RHS will also be (-1)0 =1 which is equal to LHS
Now let us take m=2
Which implies LHS will be equal to ${}^n{C_0} - {}^n{C_1}$ which is (1-n) and the RHS will be (-1)1(n-1)=(1-n) which is also same as LHS.
Now let us assume that this equation is satisfied till m=a, where a>0;
=> ${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^{a - 1}} \times {}^n{C_{a - 1}}\,$= ${( - 1)^{a - 1}}\dfrac{{(n - 1)(n - 2)........(n - a + 1)}}{{(a - 1)!}}$ . . . . . . . . . Eq 1
Now we just need to prove that the above equation is satisfied even for m=(a+1) so that it satisfies for all m.
Now place m= a+1 in the LHS , we get
${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^{a - 1}} \times {}^n{C_{a - 1}}\, + {( - 1)^a} \times {}^n{C_a}$
Which will be equal to ${( - 1)^{a - 1}}\dfrac{{(n - 1)(n - 2)........(n - a + 1)}}{{(a - 1)!}}$+ ${( - 1)^a} \times {}^n{C_a}$ [From the eq 1]
By taking (-1)a in common and multiplying ‘a’ on both numerator and denominator for the first term and also expanding ${}^n{C_a}$in the second term ,we get
${( - 1)^a}[ - \dfrac{{a(n - 1)(n - 2)...(n - a + 1)}}{{a!}} + \dfrac{{n(n - 1)(n - 2)...(n - a + 1)}}{{a!}}]$
By taking $(n - 1)(n - 2)...(n - a + 1)$as common from both terms in the above equation we get,
${( - 1)^a}(n - 1)(n - 2)...(n - a + 1)[\dfrac{{(n - a)}}{{a!}}]$
= ${( - 1)^{a - 1}}\dfrac{{(n - 1)(n - 2)........(n - a)}}{{(a)!}}$, which is RHS when placed m=a+1 in eq 1
Hence proved.

Note:
Do not make calculation mistakes while solving the induction method. You can solve this using this total simplification or you can just remember this below formula :
${}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^r} \times {}^n{C_r} = {( - 1)^r} \times {}^{n - 1}{C_{r - 1}}\,$ The asked proof is the same as the above formula when kept r as (m-1).