Question

Let \(a_1,a_2,….\) be integers such that \(a_1-a_2+a_3-a_4+…..+(-1)^{n-1}a_n=n\), for all \(n≥1\),then \(a_{51}+a_{52}+….+a_{1023} \) equals

• A. -1
• B. 10
• C. 0
• D. 1

Answer 1

The Correct Option is D

We observe a simple pattern in the values of the sequence \( a_n \): 

• If \( n \) is odd, then \( a_n = 1 \)
• If \( n \) is even, then \( a_n = -1 \)

Initial Verification

Let’s check a few initial values based on the alternating sum:

• \( n = 1 \Rightarrow a_1 = 1 \)
• \( n = 2 \Rightarrow a_1 - a_2 = 2 \Rightarrow a_2 = -1 \)
• \( n = 3 \Rightarrow a_1 - a_2 + a_3 = 3 \Rightarrow a_3 = 1 \)
• \( n = 4 \Rightarrow a_1 - a_2 + a_3 - a_4 = 4 \Rightarrow a_4 = -1 \)

This confirms the pattern: \( a_n = (-1)^{n+1} \)

Now compute:

\[ \sum_{k=51}^{1022} a_k \]

We count how many terms are there from 51 to 1022:

\[ 1022 - 51 + 1 = 972 \text{ terms} \]

Since this segment has an equal number of odd and even values:

• Odd indices \( \Rightarrow a_k = 1 \)
• Even indices \( \Rightarrow a_k = -1 \)

So, the sum becomes:

\[ \text{Sum} = \frac{972}{2} \cdot 1 + \frac{972}{2} \cdot (-1) = 0 \]

Final Step

We now evaluate the next term in the sequence:

\[ a_{1023} = 1 \quad \text{(since 1023 is odd)} \]

✅ Final Answer: \( \boxed{1} \)