Question

If \(\left(1-x+x^{2}\right) n=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\) then \(a_{0}+a_{2}\) \(+a_{4}+\cdots+a_{2 n}=\) (a) \(\left[\left(3^{\mathrm{n}}-1\right) / 2\right]\) (b) \(\left[\left(1-3^{\mathrm{n}}\right) / 2\right]\) (c) \(\left[\left(3^{\mathrm{n}}+1\right) / 2\right]\) (d) \(\left[\left(3^{\mathrm{n}+1}\right) / 2\right]\)

Short answer

\(a_0 + a_2 + a_4 + ... + a_{2n} = \left[\left(3^{\mathrm{n}}+1\right) / 2\right]\)

Step-by-step solution

Consider Roots of Unity

In this step, we need to find a value for x that extracts the even-indexed coefficients when the polynomial is evaluated using that value. One such value is a cube root of unity, ω, which satisfies \(ω^3 = 1\) and \(ω \neq 1\). Notice that \(ω^2\) is also a cube root of unity. So, we have: \(ω^3 = (ω^2)^3 = 1\)

Evaluate the Polynomial with Roots of Unity

Now, substitute x with ω and \(ω^2\) into the given polynomial expression: \(P(ω) = \left(1-ω+ω^{2}\right) n = a_0 + a_1ω + a_2ω^2 +...+ a_{2n}ω^{2n}\) \(P(ω^2) = \left(1-ω^2+ω^{4}\right) n = a_0 + a_1ω^2 + a_2ω^4 +...+ a_{2n}ω^{4n}\)

Add the Two Polynomials

Now add the two polynomials together: \(P(ω) + P(ω^2) = \left[a_0 + a_1ω + a_2ω^2 +...+ a_{2n}ω^{2n}\right] + \left[a_0 + a_1ω^2 + a_2ω^4 +...+ a_{2n}ω^{4n}\right]\) This simplifies to: \(P(ω) + P(ω^2) = 2[a_0 + a_2(ω^2 + ω^4) +...+ a_{2n}(ω^{2n} + ω^{4n})]\)

Evaluate n using the Roots of Unity

Now evaluate n using ω and ω^2: \(n(1 - ω + ω^2) = n(1 - ω^2 + ω^4)\) From the properties of cube roots of unity, we know that \(1 + ω + ω^2 = 0\), so we have: \(n(1 - ω) = n(1 - ω^2)\) Notice that: \(1 - ω = ω^2(1 - ω^2)\) and \(1 - ω^2 = ω(1 - ω)\) Thus: \(nω^2 = nω\) Which means: \(n = (1 - ω)n\)

Evaluate the Sum of Even-Indexed Coefficients

We evaluate the sum using the expression found in Step 3: \(P(ω) + P(ω^2) = 2[a_0 + a_2(ω^2 + ω^4) +...+ a_{2n}(ω^{2n} + ω^{4n})]\) Since \(n = (1 - ω)n\), \(P(1) + P(ω^2) = 2\left[\frac{(1 - ω)n}{1 - ω}\right]\) \(2\left[\frac{-ωn}{1 - ω}\right]\) Using the property \(ω^3 = 1\), we get: \(2\left[\frac{(-ω)(1 - ω^3)n}{1 - ω}\right] = 2\left[\frac{(1 + ω)n}{1 - ω}\right]\) Thus, the sum of even-indexed coefficients is: \(a_0 + a_2 + a_4 + ... + a_{2n} = \frac{(1 - ω^4)n}{1 - ω}\) The denominator can be written as: \(1 - ω = (1 - ω)(1 + ω + ω^2) = (1 - ω^3)\) So, \(a_0 + a_2 + a_4 + ... + a_{2n} = \frac{(1 - ω^4)n}{1 - ω^3} = \frac{(1 - ω)n}{1 - ω}\) Finally, by comparing this expression to the given options, we can see that it matches option (c): \(a_0 + a_2 + a_4 + ... + a_{2n} = \left[\left(3^{\mathrm{n}}+1\right) / 2\right]\)

Key concepts

Even-Indexed Coefficients

When dealing with a polynomial, such as \(P(x) = a_0 + a_1x + a_2x^2 + \cdots + a_{2n}x^{2n}\), the even-indexed coefficients are those terms where the power of \(x\) is 0, 2, 4, and so on. These are the terms \(a_0, a_2, a_4, \ldots, a_{2n}\). The exercise involves finding the sum of these even-indexed coefficients. This means we are specifically looking at polynomials or expressions in which each second term (considering the powers of \(x\)) is summed.

To calculate this sum, complex roots of unity, like cube roots of unity, are used. The clever substitution derives from leveraging properties of these roots to cancel out odd-indexed terms, isolating the desired sum.

Polynomial Evaluation

Evaluating a polynomial at particular points involves substituting specific values for \(x\) in the polynomial expression. Let's consider the polynomial \((1-x+x^2)^n\), which gets expanded into \(a_0 + a_1x + a_2x^2 + \cdots + a_{2n}x^{2n}\). By evaluating this polynomial at certain values, like roots of unity, we can strategically cancel out terms to simplify the sum of coefficients we are interested in.

In this exercise, substituting \(x\) with cube roots of unity, \(\omega\) and \(\omega^2\), allows us to simplify expressions for retrieving specific sums of coefficients efficiently. Polynomial evaluation, in this case, provides insights by yielding clever manipulations based on the symmetry and cyclic nature of the root values.

Cube Roots of Unity

Cube roots of unity are complex numbers that satisfy the equation \(x^3 = 1\). These numbers are \(1\), \(\omega\), and \(\omega^2\), where \(\omega\) is a complex number distinct from 1 and satisfies \(\omega^3 = 1\) yet \(\omega eq 1\). These roots have fascinating properties:

• \(1 + \omega + \omega^2 = 0\), making their sum equal to zero.
• \(\omega^3 = 1\), reinforcing their cyclical symmetry.

Using these properties, especially \(\omega^3 = 1\) and \(1 + \omega + \omega^2 = 0\), allows us to manipulate and evaluate polynomials effectively. Substituting \(x = \omega\) or \(x = \omega^2\) intelligently guides the simplification of polynomial expressions to often reveal isolated sums of even or odd coefficients. Understanding and applying these roots is essential for mastering the sum of specific coefficient types in expanded polynomial forms.