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

The pattern suggests that for each odd term, \(a_n​=1\), and for each even term, \(a_n​=−1\).

Therefore, for \(n=1, a_1​=1.\)

For \(n=2, a_1​−a_2​=2\), which implies \(a_2​=−1.\)

For \(n=3, a_1​−a_2​+a_3​=3,\) which implies \(a_3​=1.\)

For \(n=4, a_1​−a_2​+a_3​−a_4​=4,\) which implies \(a_4​=−1.\)

This pattern continues, with each odd term being 1 and each even term being -1.

Given this pattern, \(a_{51}​+a_{52}​+…+a_{1022}​=0\), as there are an equal number of 1's and -1's.

Therefore, the value \(a_{1023}​=1\), as it is an odd term in the sequence.