Question
Question: In the expansion of \[{\left( {1 + x} \right)^n}\], the sum of the coefficients of odd powers of \[x...
In the expansion of (1+x)n, the sum of the coefficients of odd powers of x is?
A. 2n+1
B. 2n−1
C. 2n
D. 2n−1
Solution
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,
(1+x)n
The expression (1+x)n can be expanded using the binomial theorem.
(1+x)n=nC0+nC1x+nC2x2+nC3x3+...+nCnxn
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;
(1+1)n=2(nC1+nC3+...+nCn)
Adding the left-hand side and bringing the numeric value from the right-hand side to the left-hand side, we get;
22n=(nC1+nC3+...+nCn)
Now, using the formula of exponents, we get;
nC1+nC2+nC5+...=2n−1
The correct option is D.
Note: The expansion of (1+x)n is given as the above by following the derivation given below;
(1+x)n can be written as
1+1!nx+2!n(n−1)x2+3!n(n−1)(n−2)x3+...
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 ∣x∣<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.