Question

If \[{\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .....{C_n}{x^n}\], then \[{C_1} + {C_3} + {C_5}......... = ?\]
A) \[\dfrac{{{{\left( {1 + i} \right)}^n} - {{\left( {1 - i} \right)}^n} + {2^n}}}{2}\]
B) \[\dfrac{{{{\left( {1 + i} \right)}^n} - {{\left( {1 - i} \right)}^n} + {2^n}}}{4}\]
C) \[\dfrac{{{{\left( {1 + i} \right)}^n} + {{\left( {1 - i} \right)}^n} + {2^n}}}{2}\]
D) \[\dfrac{{{{\left( {1 + i} \right)}^n} + {{\left( {1 - i} \right)}^n} + {2^n}}}{4}\]

Answer 1

Hint:
Here, we will use the concept of binomial theorem and its expansion. Then we will find the sum of binomial coefficients of the odd powers of the variable. The Binomial Theorem is the method of expanding an expression which has been raised to any finite power.

Complete Step By Step Solution
We are given with a binomial expansion.
\[ \Rightarrow {\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ........ + {C_n}{x^n}\]………………………………………. \[\left( 1 \right)\]
Substituting \[x = 1\] in the above equation, we will get
\[ \Rightarrow {\left( {1 + 1} \right)^n} = {C_0} + {C_1}(1) + {C_2}{(1)^2} + ........ + {C_n}{(1)^n}\]
We know that when 1 is raised to any power, we have 1 itself i.e., \[1^\circ = 1,{1^1} = 1,{1^2} = 1,{.......1^n} = 1\]
Adding the terms, we have
\[ \Rightarrow {\left( 2 \right)^n} = {C_0} + {C_1} + ........ + {C_n}\]
\[ \Rightarrow {C_0} + {C_1} + ........ + {C_n} = {2^n}\]…………………………………..\[\left( 2 \right)\]
Also, Rewriting equation \[\left( 1 \right)\], we get
\[ \Rightarrow - {\left( {1 + x} \right)^n} = - \left( {{C_0} + {C_1}x + {C_2}{x^2} + ........ + {C_n}{x^n}} \right)\]
Substituting \[x = - 1\] in the above equation, we get
\[ \Rightarrow - {\left( {1 - 1} \right)^n} = - \left( {{C_0} + {C_1}( - 1) + {C_2}{{( - 1)}^2} + ........ + {C_n}{{( - 1)}^n}} \right)\]
We know that when \[ - 1\] is raised to odd power, we get \[ - 1\]. When \[ - 1\] is raised to even power, we get 1.
i.e., \[{( - 1)^n} = - 1\] when \[n\] is an odd number; \[{( - 1)^n} = 1\]when \[n\] is an even number.
Subtracting the terms, we get
\[ \Rightarrow - {\left( 0 \right)^n} = - \left( {{C_0} - {C_1} + {C_2} - {C_3} + ...........} \right)\]
Multiplying the terms with negative sign, we get
\[ \Rightarrow 0 = - {C_0} + {C_1} - {C_2} + {C_3} - ...........\]
\[ \Rightarrow - {C_0} + {C_1} - {C_2} + {C_3} - ........... = 0\]……………………………..\[\left( 3 \right)\]
Adding equations \[\left( 2 \right)\] and equations \[\left( 3 \right)\], we get
\[ \Rightarrow {C_0} + {C_1} + {C_2} + ........... - {C_0} + {C_1} - {C_2} + ....... = {2^n} + 0\]
Odd terms get subtracted as they have unlike signs and even terms get added as they have like signs.
So, we have
\[ \Rightarrow 2{C_1} + 2{C_3} + 2{C_5} + ......... = {2^n}\]
Taking out the common terms, we get
\[ \Rightarrow 2\left( {{C_1} + {C_3} + {C_5} + .........} \right) = {2^n}\]
Dividing both sides by \[2\], we get
\[ \Rightarrow \left( {{C_1} + {C_3} + {C_5} + .........} \right) = \dfrac{{{2^n}}}{2}\]
\[ \Rightarrow \left( {{C_1} + {C_3} + {C_5} + .........} \right) = {2^{n - 1}}\]
We know that \[{\left( {1 + i} \right)^n} + {\left( {1 - i} \right)^n} = {\left( {\sqrt 2 } \right)^{n + 2}}\cos \dfrac{{n\pi }}{4}\] .
Since we have to find the sum of binomial coefficients of odd powers, we will take \[\cos \dfrac{{n\pi }}{4} = \dfrac{1}{{\sqrt 2 }}\].
So, we get \[\dfrac{{{{\left( {1 + i} \right)}^n} + {{\left( {1 - i} \right)}^n} + {2^n}}}{4} = {2^{n - 1}}\]
Therefore, \[{C_1} + {C_3} + {C_5}......... = \dfrac{{{{\left( {1 + i} \right)}^n} + {{\left( {1 - i} \right)}^n} + {2^n}}}{4}\]

Hence, option D is the correct answer.

Note:
The binomial coefficient uses the concept of Combinations. Binomial coefficients refer to the integers which are coefficients in the binomial theorem. Binomial theorem has a wide range of applications in mathematics like finding the remainder, finding digits of a number, etc. A combination is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of the selection does not matter.