If C₀, C₁, C₂, , Cₙ are the binomial coefficients in the expansion of (1+x)ⁿ, then (C₀+C₁) - (C₂+C₃) + (C₄+C₅) - (C₆+C₇) + ·s = [AP EAPCET 23 May 2025 S2]

Correct Answer is : 2n2(cos⁡nπ4+sin⁡nπ4)\; 2ⁿ/2} ≤ft( nπ/4 + nπ/4 )22n(cos4nπ+sin4nπ)

\; 2ⁿ/2} ≤ft( nπ/4 + i nπ/4 )

\; 2ⁿ/2} ≤ft( nπ/3 + nπ/3 )

\; 2ⁿ/2} ≤ft( nπ/3 + i nπ/3 )

\; 2ⁿ/2} ≤ft( nπ/4 + nπ/4 )

Correct Answer is : 2n2(cos⁡nπ4+sin⁡nπ4)\; 2ⁿ/2} ≤ft( nπ/4 + nπ/4 )22n(cos4nπ+sin4nπ)

Test Index

Binomial Theorem

Section: Mathematics

Question : 6 of 33

Marks: +1, -0

C_0, C_1, C_2, \ldots, C_n

(1+x)^n

(C_0+C_1) - (C_2+C_3) + (C_4+C_5) - (C_6+C_7) + \cdots =

Solution:

\because (1+x)^n=\sum\limits_{k=0}^{n} C_k x^k

and

(C_0+C_1) - (C_2+C_3) + (C_4+C_5) - \cdots

can be grouped as

\sum\limits_{k=0}^{\frac{n}{2}} (-1)^k (C_{2k}+C_{2k+1})

\; \text{ and } \; (1+i)^n=\sum\limits_{k=0}^{n} C_k i^k

\operatorname{Re}[(1+i)^n] = C_0 - C_2 + C_4 - C_6 + \cdots

\operatorname{Im}[(1+i)^n] = C_1 - C_3 + C_5 - C_7 + \cdots

Adding both

\; \operatorname{Re}[(1+i)^n] + \operatorname{Im}[(1+i)^n]

\; = (C_0+C_1) - (C_2+C_3) + (C_4+C_5) - \ldots

Now,

(1+i)^n

= (\sqrt{2})^n \left( \cos\left(\frac{n\pi}{4}\right) + i \sin\left(\frac{n\pi}{4}\right) \right)

Hence,

(C_0+C_1) - (C_2+C_3) + (C_4+C_5)

- \ldots \; = \operatorname{Re}[(1+i)^n] + \operatorname{Im}[(1+i)^n]

\; = 2^{\frac{n}{2}} \left( \cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{4}\right) \right)

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