Question:
Let $a_{1}, a_{2}, \ldots$ be integers such that
$a_{1}-a_{2}+a_{3}-a_{4}+\cdots+(-1)^{n-1} a_{n}=n,$ for all $n \geq 1$
Then $a_{51}+a_{52}+\cdots+a_{1023}$ equals
- -1
- 10
- 0
- 1
Correct Answer: Option: 4
for $n=1, \quad a_{1}=n \Rightarrow a_{1}=1$
for $n=2, \quad a_{1}-a_{2}=2 \Rightarrow a_{2}=-1$
for $n=3, \quad a_{1}-a_{2}+a_{3}=3 \Rightarrow a_{3}=1$
for $n=4, \quad a_{1}-a_{2}+a_{3}-a_{4}=4 \Rightarrow a_{4}=-1$
From the pattern, each odd term = 1 and each even term = -1
$\Rightarrow a_{51}+a_{52}+\cdots+a_{1022}=0$
Therefore the value is equal to $a_{1023}=1$
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