Sum of terms in a sequence

Let $S_n = a_1 - a_2 + a_3 - \dots + (-1)^{n-1} a_n = n$.

For $n=1$: $a_1 = 1$. For $n=2$: $a_1 - a_2 = 2 \implies 1 - a_2 = 2 \implies a_2 = -1$. For $n=3$: $a_1 - a_2 + a_3 = 3 \implies 2 + a_3 = 3 \implies a_3 = 1$. For $n=4$: $a_1 - a_2 + a_3 - a_4 = 4 \implies 3 - a_4 = 4 \implies a_4 = -1$.

Thus, $a_n = 1$ when $n$ is odd, and $a_n = -1$ when $n$ is even.

We need to find $a_{51} + a_{52} + \dots + a_{1023}$. The total number of terms is $1023 - 51 + 1 = 973$. The terms alternate: $1, -1, 1, -1, \dots, 1$. Since there are 487 pairs of $(1, -1)$ which sum to $0$, and one additional term $a_{1023} = 1$, the total sum is $1$.