Consider a sequence of real number $x_1, x_2, x_3, ...$ such that $x_{n+1} = x_n + n - 1$ for all $n \ge 1$. If $x_1 = -1$ then $x_{100}$ is equal to

We have: $x_{n+1} = x_n + n - 1$ for all $n \geq 1$, with $x_1 = -1$

This means each term equals the previous term plus $(n-1)$.

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Let's calculate the first few terms:

$x_1 = -1$ (given)

$x_2 = x_1 + (1-1) = -1 + 0 = -1$

$x_3 = x_2 + (2-1) = -1 + 1 = 0$

$x_4 = x_3 + (3-1) = 0 + 2 = 2$

$x_5 = x_4 + (4-1) = 2 + 3 = 5$

$x_6 = x_5 + (5-1) = 5 + 4 = 9$

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Key Insight: Instead of trying to find a pattern in the terms themselves, let's look at the differences between consecutive terms.

From our recurrence relation: $x_{n+1} - x_n = n - 1$

This gives us:

• $x_2 - x_1 = 1 - 1 = 0$
• $x_3 - x_2 = 2 - 1 = 1$
• $x_4 - x_3 = 3 - 1 = 2$
• $x_5 - x_4 = 4 - 1 = 3$
• $\vdots$
• $x_n - x_{n-1} = (n-1) - 1 = n - 2$

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When we add all these differences from $x_2$ to $x_n$:

$(x_2 - x_1) + (x_3 - x_2) + (x_4 - x_3) + \cdots + (x_n - x_{n-1}) = x_n - x_1$

All the middle terms cancel out! This is called a telescoping sum.

So: $x_n - x_1 = 0 + 1 + 2 + 3 + \cdots + (n-2)$

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The sum $0 + 1 + 2 + 3 + \cdots + (n-2)$ is the sum of first $(n-2)$ natural numbers (including 0).

For our sum: $0 + 1 + 2 + \cdots + (n-2) = \dfrac{(n-2)(n-1)}{2}$

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$x_n - x_1 = \dfrac{(n-2)(n-1)}{2}$

Since $x_1 = -1$:

$x_n = -1 + \dfrac{(n-2)(n-1)}{2}$

$x_n = \dfrac{-2 + (n-2)(n-1)}{2}$

$x_n = \dfrac{(n-2)(n-1) - 2}{2}$

$x_n = \dfrac{n^2 - 3n + 2 - 2}{2}$

$x_n = \dfrac{n^2 - 3n}{2} = \dfrac{n(n-3)}{2}$

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Let's check our formula with known values:

$x_4 = \dfrac{4(4-3)}{2} = \dfrac{4 \times 1}{2} = 2$

$x_5 = \dfrac{5(5-3)}{2} = \dfrac{5 \times 2}{2} = 5$

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$x_{100} = \dfrac{100(100-3)}{2} = \dfrac{100 \times 97}{2} = \dfrac{9700}{2} = 4850$

Therefore, $x_{100} = 4850$