Question

If \(n\in Z\), then the value of \(\frac{2^{n}}{(1+i)^{2n}}+\frac{2^{n}}{(1-i)^{2n}}+\frac{(1+i)^{2n}+(1-i)^{2n}}{2^{n}}\) is equal to

• A. \(4\,\cos \left(\frac{n\pi }{2}\right)\)
• B. \(2(n+1)\cos \left(\frac{n\pi }{2}\right)\)
• C. \(2[1+(-1)^{n}]\)
• D. \(2(n+2)\)

Hint:

Use (1+i)^2 = 2i and (1-i)^2 = -2i, then powers of i.

Answer 1

The Correct Option is A

Step 1: Test n = 0
Each of the first two terms is 1 and the last is 2, so the value is 4. Option $4\cos 0=4$ matches.

Step 2: Test n = 1
The first two terms are $-i$ and $i$, and the last is $i-i=0$, so the total is 0, matching $4\cos\dfrac{\pi}{2}=0$.

Step 3: Eliminate
The other options give non-zero values at n = 1, so they fail.

Final Answer:
The values at n = 0 and n = 1 match 4 cos(n pi by 2) only. \[ \boxed{\text{(A) }4\cos\left(\dfrac{n\pi}{2}\right)} \]