Question

The sum of the series ${}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + ........................{}^{20}{C_{10}}$ is:
1) $\dfrac{1}{2}{}^{20}{C_{10}}$
2) ${}^{20}{C_{10}}$
3) $ - {}^{20}{C_{10}}$
4) 0

Answer 1

Hint: The above problem is based on the Binomial theorem concept;
Binomial theorem: As the power in an expansion increases calculation becomes lengthy and tedious to solve, then to ease out the calculation we use the Binomial theorem concept.
A general expression of Binomial theorem is:
\[{(a + b)^n} = \sum\limits_{k = 0}^n {\left( {\frac{n}{k}} \right)} {a^{(n - k)}}{b^k}\],
where k and n are the limits of the summation, a and b are the constants .
Using the above theorem we will find out the expansion of the series given in the question.

Complete step-by-step answer:
In order to solve the above problem first we will understand the Binomial theorem in detail and then we will proceed for the calculation part of the problem.
The Binomial theorem describes the algebraic expansion of powers of a binomial.
According to the Binomial theorem, it is possible to expand the polynomial (x + y)n into a sum involving the terms of the form a${x^b}$${y^c}$, where the exponents b and c are nonnegative integers with b + c =n, and the coefficient a of each term is a specific positive integer depending on n and b.
Now we will do the calculation part of the problem;
$ \Rightarrow$ ${\left( {1 + x} \right)^{20}} = {}^{20}{C_0} + {}^{20}{C_1}x + {}^{20}{C_2}{x^2} + {}^{20}{C_3}{x^3}...............{}^{20}{C_{20}}{x^{20}}$ (We have expanded the expression (1+x) raise to power 20 as per binomial theorem, where is the C is the variable of combination (selection of items from a collection))
From the above expression we replace $x = - 1$,
$ \Rightarrow {\left( {1 - 1} \right)^{20}} = {}^{20}{C_0} + {}^{20}{C_1}( - 1) + {}^{20}{C_2}{( - 1)^2} + {}^{20}{C_3}{( - 1)^3}...............{}^{20}{C_{20}}{( - 1)^{20}}$
$ \Rightarrow 0 = {}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3}............... + {}^{20}{C_{20}}$( LHS becomes zero and odd power gives negative sign)
$ \Rightarrow 0 = {}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + {}^{20}{C_9} + {}^{20}{C_{10}} + {}^{20}{C_9} + {}^{20}{C_8}........... + {}^{20}{C_0}$(Because ${}^n{C_r} = {}^n{C_{n - r}}$ )
Now in the next expression we will group the terms which are being repeated twice in the series
$ \Rightarrow 0 = 2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + {}^{20}{C_4}........ - {}^{20}{C_9}) + {}^{20}{C_{10}}$
$ \Rightarrow - {}^{20}{C_{10}} = 2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + {}^{20}{C_4}........ - {}^{20}{C_9})$
On adding 2${}^{20}{C_{10}}$ on both LSH and RHS
$ \Rightarrow - {}^{20}{C_{10}} + 2{}^{20}{C_{10}} = 2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + {}^{20}{C_4}........ - {}^{20}{C_9} + {}^{20}{C_{10}})$
$ \Rightarrow {}^{20}{C_{10}} = 2({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + {}^{20}{C_4}........ - {}^{20}{C_9} + {}^{20}{C_{10}})$
$ \Rightarrow \dfrac{1}{2}{}^{20}{C_{10}} = ({}^{20}{C_0} - {}^{20}{C_1} + {}^{20}{C_2} - {}^{20}{C_3} + {}^{20}{C_4}........ - {}^{20}{C_9} + {}^{20}{C_{10}})$(We have shifted the 2 from RHS to LHS )

Thus, option 1 is correct.

Note: Binomial theorem and its generalization can be used to prove results and solve problems in combinations, algebra, calculus and many areas of mathematics. Binomial theorem helps in exploring probability in an organized way and statistical calculations as well.