Question

The value of ${}^{40}{C_0} + {}^{40}{C_1} + {}^{40}{C_2} + ... + {}^{40}{C_{20}}$ is ?
A. ${2^{40}} + \dfrac{{40!}}{{{{(20!)}^2}}}$
B. ${2^{39}} - \dfrac{1}{2} \times \dfrac{{40!}}{{{{(20!)}^2}}}$
C. ${2^{39}} + {}^{40}{C_{20}}$
D. None of these

Answer 1

Here in this question we have to solve the given binomial equation. By using the definitions and properties of the binomial theorem we simplify the given equation. While solving the above equation we use the simple arithmetic operations and hence we obtain the required solution for the question.

Complete step by step solution:
We know the formula ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ .
Binomial theorem, statement that for any positive integer n, the nth power of the sum of two numbers a and b may be expressed as the sum of n + 1 terms of the form.
We also use the properties and definitions of ${}^n{C_r}$ . We simplify the given equation.
Now consider the given binomial equation
${}^{40}{C_0} + {}^{40}{C_1} + {}^{40}{C_2} + ... + {}^{40}{C_{20}}$
divide and multiply the equation by 2 we have
$\Rightarrow \dfrac{1}{2} \times 2\left[ {{}^{40}{C_0} + {}^{40}{C_1} + {}^{40}{C_2} + ... + {}^{40}{C_{20}}} \right]$
On multiplying we get
$\Rightarrow \dfrac{1}{2}\left[ {2.{}^{40}{C_0} + 2.{}^{40}{C_1} + 2.{}^{40}{C_2} + ... + 2.{}^{40}{C_{20}}} \right]$
By the property of combinations ${}^n{C_r} = {}^n{C_{n - r}}$, using this the above equation is written as
$\Rightarrow \dfrac{1}{2}\left[ {({}^{40}{C_0} + {}^{40}{C_{40}}) + ({}^{40}{C_1} + {}^{40}{C_{39}}) + ...{(^{40}}{C_{19}} + {}^{40}{C_{21}}) + 2.{}^{40}{C_{20}}} \right]$
so it can be rewritten as
$\Rightarrow \dfrac{1}{2}\left[ {({}^{40}{C_0} + {}^{40}{C_1} + {}^{40}{C_2} + {}^{40}{C_3} + ...{}^{40}{C_{20}} + ..{ + ^{40}}{C_{39}} + {}^{40}{C_{40}}) + {}^{40}{C_{20}}} \right]$
By the property of combinations ${}^n{C_1} + {}^n{C_2} + {}^n{C_3} + ... + {}^n{C_n} = {2^n}$, using this the above equation is written as
$\Rightarrow \dfrac{1}{2}\left[ {{2^{40}} + {}^{40}{C_{20}}} \right]$
This can be written on further simplification is
$\Rightarrow {2^{39}} + \dfrac{1}{2}{}^{40}{C_{20}}$
Hence we have simplified the given binomial equation and determined the value. but the obtained is not matching to anyone of the options.
Therefore the option D is the correct one.

So, the correct answer is “Option D”.

Note: The binomial expansion formula and the formula is defined as ${(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ... + {}^n{C_n}{a^0}{b^n}$ by substituting the value of a, b and n we can calculate the solution for the expansion. We know the formula ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ . We cannot apply this, Because it is time consuming and a very long procedure.