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

Answer: (D)

1

Answer 2

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.