If (1+x)^n = C0 + C1x + C2x^2 + ... + Cnx^n, then C0 - C2 + C4 - C6 + ... =

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – AP-EAMCET 2026

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Home Maths Binomial TheoremQuestion #34

(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}

, then

C_{0} - C_{2} + C_{4} - C_{6} + \dots =

Clue 1 of 3: Binomial Theorem and Complex Numbers

The coefficients $C_{k}$ are from the binomial expansion. The alternating sum involving even coefficients suggests using complex numbers.

🧠 Full Solution Path

Step 1: Recall the Binomial Expansion and Substitute a Complex Value✦ Active

The given binomial expansion is $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + C_{3} x^{3} + C_{4} x^{4} + \dots + C_{n} x^{n}$ . To obtain the alternating sum of even coefficients ( $C_{0} - C_{2} + C_{4} - \dots$ ), we should substitute a value of $x$ such that powers of $x$ cycle through $1, i, - 1, - i$ . The complex number $x = i$ (where $i = \sqrt{- 1}$ ) is suitable for this.

(1 + i)^{n} = C_{0} + C_{1} i + C_{2} i^{2} + C_{3} i^{3} + C_{4} i^{4} + C_{5} i^{5} + \dots

Using the properties of powers of $i$ ( $i^{2} = - 1$ , $i^{3} = - i$ , $i^{4} = 1$ , $i^{5} = i$ , etc.):

(1 + i)^{n} = C_{0} + C_{1} i - C_{2} - C_{3} i + C_{4} + C_{5} 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} - \dots)

The required expression is the real part of $(1 + i)^{n}$ .

💡 Teacher's Secret Hint

Remember that $i^{2} = - 1$ , $i^{3} = - i$ , and $i^{4} = 1$ . This cyclical property of powers of $i$ is key to forming the alternating sum of coefficients.

Step 2: Convert

(1 + i)

to Polar Form○ Expand

To evaluate $(1 + i)^{n}$ , convert the complex number $1 + i$ into its polar form $r (\cos θ + i \sin θ)$ .

The modulus $r = | 1 + i | = \sqrt{(Re)^{2} + (Im)^{2}} = \sqrt{1^{2} + 1^{2}} = \sqrt{2}$ .

The argument $θ = \arg (1 + i) = \arctan (\frac{1}{1}) = \frac{π}{4}$ (since $1 + i$ is in the first quadrant).

Thus, $1 + i$ in polar form is:

1 + i = \sqrt{2} (\cos \frac{π}{4} + i \sin \frac{π}{4})

💡 Teacher's Secret Hint

Always check the quadrant of the complex number when finding the argument to ensure the correct angle is chosen.

Step 3: Apply De Moivre's Theorem○ Expand

Now, raise the polar form to the power $n$ using De Moivre's Theorem, which states that $(r (\cos θ + i \sin θ))^{n} = r^{n} (\cos (n θ) + i \sin (n θ))$ .

(1 + i)^{n} = {(\sqrt{2} (\cos \frac{π}{4} + i \sin \frac{π}{4}))}^{n}

(1 + i)^{n} = (\sqrt{2})^{n} (\cos (\frac{n π}{4}) + i \sin (\frac{n π}{4}))

Since $\sqrt{2} = 2^{1 / 2}$ , we have $(\sqrt{2})^{n} = (2^{1 / 2})^{n} = 2^{n / 2}$ . Therefore:

(1 + i)^{n} = 2^{n / 2} \cos (\frac{n π}{4}) + i \cdot 2^{n / 2} \sin (\frac{n π}{4})

💡 Teacher's Secret Hint

De Moivre's theorem simplifies raising complex numbers to powers significantly. Ensure you distribute the power to both the modulus and the argument.

Step 4: Equate Real Parts○ Expand

Comparing the expression for $(1 + i)^{n}$ from Step 1 and Step 3:

(C_{0} - C_{2} + C_{4} - C_{6} + \dots) + i (C_{1} - C_{3} + C_{5} - \dots) = 2^{n / 2} \cos (\frac{n π}{4}) + i \cdot 2^{n / 2} \sin (\frac{n π}{4})

Equating the real parts on both sides, we get:

C_{0} - C_{2} + C_{4} - C_{6} + \dots = 2^{n / 2} \cos (\frac{n π}{4})

This matches option 1.

💡 Teacher's Secret Hint

When comparing complex numbers, remember that their real parts must be equal, and their imaginary parts must also be equal.

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Maths Question 9 – AP-EAMCET 2026

Clue 1 of 3: Binomial Theorem and Complex Numbers

🧠 Full Solution Path