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 $n \ge 1$. Then $a_{51} + a_{52} + ... + a_{1023}$ equals

We need to find the pattern of the sequence $a_1, a_2, a_3, ...$ and then calculate the sum from $a_{51}$ to $a_{1023}$.

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Given: $a_1 - a_2 + a_3 - a_4 + ... + (-1)^{n-1} a_n = n$ for all $n \geq 1$

Let us call this sum $S_n = n$. So:

$S_1 = a_1 = 1$

$S_2 = a_1 - a_2 = 2$

$S_3 = a_1 - a_2 + a_3 = 3$

$S_4 = a_1 - a_2 + a_3 - a_4 = 4$

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Key Insight: Notice that $S_{n+1} - S_n = 1$ for all $n \geq 1$.

This means: $S_{n+1} - S_n = (-1)^n a_{n+1} = 1$

Therefore: $a_{n+1} = \frac{1}{(-1)^n} = (-1)^n$

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Using $a_{n+1} = (-1)^n$:

$a_2 = (-1)^1 = -1$

$a_3 = (-1)^2 = 1$

$a_4 = (-1)^3 = -1$

$a_5 = (-1)^4 = 1$

Pattern Discovery: $a_n = (-1)^{n-1}$ for all $n \geq 1$

Why this works: For any integer $n$, if $n$ is odd, then $n-1$ is even, so $(-1)^{n-1} = 1$. If $n$ is even, then $n-1$ is odd, so $(-1)^{n-1} = -1$.

Therefore:

$a_{\text{odd}} = 1$

$a_{\text{even}} = -1$

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We need: $a_{51} + a_{52} + ... + a_{1023}$

Since our pattern is $a_{\text{odd}} = 1$ and $a_{\text{even}} = -1$:

$a_{51} = 1$ (51 is odd)

$a_{52} = -1$ (52 is even)

$a_{53} = 1$ (53 is odd)

And so on...

The sum becomes: $1 + (-1) + 1 + (-1) + ... + 1$

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Total terms from 51 to 1023: $1023 - 51 + 1 = 973$ terms

Since we start with $a_{51} = 1$ and the pattern alternates, we have consecutive pairs $(1, -1)$ that sum to 0.

Number of complete pairs: $\lfloor\frac{973}{2}\rfloor = 486$ pairs

Each pair contributes 0 to the sum.

Remaining terms: $973 - 2(486) = 1$ term

This remaining term is $a_{1023} = 1$ (since 1023 is odd).

Therefore: $a_{51} + a_{52} + ... + a_{1023} = 486 \times 0 + 1 = 1$