Question

If \(C_0, C_1, C_2, \dots, C_n\) are the binomial coefficients in the expansion of \((1 + x)^n\), then \((C_0 + C_1) - (C_2 + C_3) + (C_4 + C_5) - (C_6 + C_7) + \dots = \)

• A. \(2^{n/2} \left(\cos\left(\frac{n\pi}{4}\right) + i\sin\left(\frac{n\pi}{4}\right)\right)\)
• B. \(2^{n/2} \left(\cos\left(\frac{n\pi}{3}\right) + \sin\left(\frac{n\pi}{3}\right)\right)\)
• C. \(2^{n/2} \left(\cos\left(\frac{n\pi}{3}\right) + i\sin\left(\frac{n\pi}{3}\right)\right)\)
• D. \(2^{n/2} \left(\cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{4}\right)\right)\)

Hint:

Sums involving alternating binomial coefficients can often be evaluated by considering complex numbers. Substitute \(x=i\) (or \(x=-i\)) into the binomial expansion \((1+x)^n\) and then convert the complex number to polar form to relate the real and imaginary parts to sums of coefficients.

Answer 1

The Correct Option is D

Step 1: Write the binomial expansion of \((1+x)^n\).
The binomial expansion of \((1+x)^n\) is given by: \[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + C_3 x^3 + C_4 x^4 + C_5 x^5 + C_6 x^6 + C_7 x^7 + \dots + C_n x^n. \] Step 2: Substitute \(x=i\) into the binomial expansion.
Let \(x=i\). Then: \[ (1+i)^n = C_0 + C_1 i + C_2 i^2 + C_3 i^3 + C_4 i^4 + C_5 i^5 + C_6 i^6 + C_7 i^7 + \dots \] Recall the powers of \(i\): \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and the pattern repeats. Substitute these values: \[ (1+i)^n = C_0 + C_1 i - C_2 - C_3 i + C_4 + C_5 i - C_6 - C_7 i + \dots \] Group the real and imaginary parts: \[ (1+i)^n = (C_0 - C_2 + C_4 - C_6 + \dots) + i(C_1 - C_3 + C_5 - C_7 + \dots). \quad (\text{Equation 1}) \] Step 3: Express \((1+i)^n\) in polar form using De Moivre's Theorem.
First, convert \(1+i\) to polar form \(re^{i\theta}\):
Magnitude \(r = |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
Argument \(\theta = \arg(1+i) = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}\).
So, \(1+i = \sqrt{2}\left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) = \sqrt{2}e^{i\frac{\pi}{4}}\). Now, raise to the power of \(n\): \[ (1+i)^n = (\sqrt{2})^n \left(\cos\left(\frac{n\pi}{4}\right) + i\sin\left(\frac{n\pi}{4}\right)\right) = 2^{n/2} \left(\cos\left(\frac{n\pi}{4}\right) + i\sin\left(\frac{n\pi}{4}\right)\right). \quad (\text{Equation 2}) \] Step 4: Equate the real and imaginary parts from Equation 1 and Equation 2.
Comparing the real parts:
\[ C_0 - C_2 + C_4 - C_6 + \dots = 2^{n/2} \cos\left(\frac{n\pi}{4}\right). \] Comparing the imaginary parts: \[ C_1 - C_3 + C_5 - C_7 + \dots = 2^{n/2} \sin\left(\frac{n\pi}{4}\right). \] Step 5: Evaluate the given expression.
The expression to be evaluated is \(S = (C_0 + C_1) - (C_2 + C_3) + (C_4 + C_5) - (C_6 + C_7) + \dots\).
Rearrange the terms: \[ S = C_0 + C_1 - C_2 - C_3 + C_4 + C_5 - C_6 - C_7 + \dots \] This can be written as the sum of the real and imaginary parts derived in Step 4: \[ S = (C_0 - C_2 + C_4 - C_6 + \dots) + (C_1 - C_3 + C_5 - C_7 + \dots). \] Substitute the expressions from Step 4: \[ S = 2^{n/2} \cos\left(\frac{n\pi}{4}\right) + 2^{n/2} \sin\left(\frac{n\pi}{4}\right). \] Factor out \(2^{n/2}\): \[ S = 2^{n/2} \left(\cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{4}\right)\right). \] This matches Option (4).