
In the expansion of \[{\left( {1 + x} \right)^n}\], the sum of the coefficients of odd powers of \[x\] is?
A. \[{2^n} + 1\]
B. \[{2^n} - 1\]
C. \[{2^n}\]
D. \[{2^{n - 1}}\]
Answer
569.1k+ views
Hint: We have to note down the given expression and expand it using the binomial expansion. Eliminate all the even terms out by adding or subtracting a certain constant which suits. Then solve it and simplify it to get the final answer.
Complete step-by-step solution:
Given expression,
\[{\left( {1 + x} \right)^n}\]
The expression \[{\left( {1 + x} \right)^n}\] can be expanded using the binomial theorem.
\[{\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + ... + {}^n{C_n}{x^n}\]
Now, substituting the value of \[x\] with \[x = 1\] and on the right hand side, subtracting \[1\] to eliminate the even values of the expression, we get;
\[{\left( {1 + 1} \right)^n} = 2\left( {{}^n{C_1} + {}^n{C_3} + ... + {}^n{C_n}} \right)\]
Adding the left-hand side and bringing the numeric value from the right-hand side to the left-hand side, we get;
\[\dfrac{{{2^n}}}{2} = \left( {{}^n{C_1} + {}^n{C_3} + ... + {}^n{C_n}} \right)\]
Now, using the formula of exponents, we get;
\[{}^n{C_1} + {}^n{C_2} + {}^n{C_5} + ... = {2^{n - 1}}\]
The correct option is D.
Note: The expansion of \[{\left( {1 + x} \right)^n}\] is given as the above by following the derivation given below;
\[{\left( {1 + x} \right)^n}\] can be written as
\[1 + \dfrac{n}{{1!}}x + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ...\]
The expression is true for all the real values of \[n\] although there are no conditions on \[x\].
If \[n\] is a positive integer, then the expansion is terminated, but if \[n\] is a negative integer or not an integer or both a combination of an integer and any other subject value, we have an infinite series which is only valid when \[\left| x \right| < 1\].
In algebra, especially in permutations and combinations, the binomial theorem describes the algebraic expansion of powers of a binomial, i.e., an expression containing two elements. It can be a polynomial, which is taken in terms of binomial by merging two elements or more into a single element and then applying the same binomial theorem to the expression.
Complete step-by-step solution:
Given expression,
\[{\left( {1 + x} \right)^n}\]
The expression \[{\left( {1 + x} \right)^n}\] can be expanded using the binomial theorem.
\[{\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + {}^n{C_3}{x^3} + ... + {}^n{C_n}{x^n}\]
Now, substituting the value of \[x\] with \[x = 1\] and on the right hand side, subtracting \[1\] to eliminate the even values of the expression, we get;
\[{\left( {1 + 1} \right)^n} = 2\left( {{}^n{C_1} + {}^n{C_3} + ... + {}^n{C_n}} \right)\]
Adding the left-hand side and bringing the numeric value from the right-hand side to the left-hand side, we get;
\[\dfrac{{{2^n}}}{2} = \left( {{}^n{C_1} + {}^n{C_3} + ... + {}^n{C_n}} \right)\]
Now, using the formula of exponents, we get;
\[{}^n{C_1} + {}^n{C_2} + {}^n{C_5} + ... = {2^{n - 1}}\]
The correct option is D.
Note: The expansion of \[{\left( {1 + x} \right)^n}\] is given as the above by following the derivation given below;
\[{\left( {1 + x} \right)^n}\] can be written as
\[1 + \dfrac{n}{{1!}}x + \dfrac{{n\left( {n - 1} \right)}}{{2!}}{x^2} + \dfrac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{{3!}}{x^3} + ...\]
The expression is true for all the real values of \[n\] although there are no conditions on \[x\].
If \[n\] is a positive integer, then the expansion is terminated, but if \[n\] is a negative integer or not an integer or both a combination of an integer and any other subject value, we have an infinite series which is only valid when \[\left| x \right| < 1\].
In algebra, especially in permutations and combinations, the binomial theorem describes the algebraic expansion of powers of a binomial, i.e., an expression containing two elements. It can be a polynomial, which is taken in terms of binomial by merging two elements or more into a single element and then applying the same binomial theorem to the expression.
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