Question

Let \( \binom{n}{r} \) denote the number of ways of choosing \( r \) distinct objects out of \( n \) distinct objects.

• A. \( \sum_{k=0}^6 (-1)^k \binom{13}{2k} = -64. \)
• B. \( \sum_{k=0}^6 (-1)^k \binom{13}{2k+1} = 128. \)
• C. \( \sum_{k=0}^6 (-1)^k \binom{13}{2k} = 64. \)
• D. \( \sum_{k=0}^6 (-1)^k \binom{13}{2k+1} = 128. \)

Hint:

When solving sums involving binomial coefficients, using known identities and simplifying the terms can help identify the correct answer.

Answer 1

The Correct Option is A

Step 1: Use the binomial expansion of $(1+i)^{13}$.
By the binomial theorem, $(1+i)^{13}=\sum_{j=0}^{13}\binom{13}{j} i^{\,j}$. The real part collects the even $j$ and the imaginary part collects the odd $j$.

Step 2: Real part gives the even sum.
Since $i^{2k}=(-1)^k$, the real part is $\sum_k (-1)^k\binom{13}{2k}$, which is exactly the sum in options A and C.

Step 3: Compute $(1+i)^{13}$.
Write $1+i=\sqrt2\,e^{i\pi/4}$. Then $(1+i)^{13}=2^{13/2}e^{i\,13\pi/4}$. Now $\frac{13\pi}{4}=3\pi+\frac{\pi}{4}$, so the angle points to $\cos,\sin$ values of $-\frac{1}{\sqrt2}$ each.

Step 4: Read off the parts.
So $(1+i)^{13}=2^{6.5}\big(-\tfrac{1}{\sqrt2}-\tfrac{i}{\sqrt2}\big)=2^{6}(-1-i)=-64-64i$.

Step 5: Match to the options.
The real part is $-64$, so the even sum equals $-64$. That is option A. The odd sum is the imaginary part, also $-64$, so the options claiming $128$ are wrong.
\[ \boxed{\text{Option A: even sum} = -64} \]