Question

The value of \[{}^n{C_0} - {}^n{C_1} + {}^n{C_2} - ...... + {( - 1)^{{n^n}}}{C_n}\]is:
   (A) 1
   (B) 0
   (c) 2
   (c) n

Answer 1

Hint- Proceed the solution with the help of binomial expansion. The binomial expansion of ${(1 - x)^n} = {}^n{C_0} + {}^n{C_1}{( - x)^1} + {}^n{C_2}{( - x)^2} + .......... + {}^n{C_n}{( - x)^n}$. If however we remove the variable x then we can get the series of binomial coefficients.

Complete step-by-step solution -
The binomial expansion of ${(1 - x)^n} = {}^n{C_0} + {}^n{C_1}{( - x)^1} + {}^n{C_2}{( - x)^2} + .......... + {}^n{C_n}{( - x)^n}$.
Put x=1 in this expansion.
Since, we know that ${(1 - x)^n} = {}^n{C_0} + {}^n{C_1}{( - x)^1} + {}^n{C_2}{( - x)^2} + .......... + {}^n{C_n}{( - x)^n}$
Putting x=1 on both sides,
${(1 - 1)^n} = {}^n{C_0} + {}^n{C_1}{( - 1)^1} + {}^n{C_2}{( - 1)^2} + .......... + {}^n{C_n}{( - 1)^n}$$ \Rightarrow 0 = {}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^n}{}^n{C_n}$
$ \Rightarrow {}^n{C_0} - {}^n{C_1} + {}^n{C_2} - .......... + {( - 1)^n}{}^n{C_n} = 0$


Note- Binomial theorem is a fast method of expanding a binomial expression that has been raised to some large power.
we got the expansion of ${(1 - x)^n}$ using the expansion of
${(1 + x)^n} = $ ${}^n{C_0} + {}^n{C_1}{(x)^1} + {}^n{C_2}{(x)^2} + .......... + {}^n{C_n}{(x)^n}$, as ${(1 - x)^n}$= ${(1 + ( - x))^n}$.
The expansion can also be written as ${(1 + x)^n} = $ $1 + \dfrac{n}{{1!}}x + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ........$