Question

Consider a sequence of real numbers \(x_1,x_2,x_3,…\) such that \(x_{n+1}=x_n+n−1\) for all \(n≥1\). If \(x_1=−1\) then \(x_{100}\) is equal to

• A. 4949
• B. 4849
• C. 4850
• D. 4950

Answer 1

The Correct Option is C

Given the recurrence relation: 
\( x_{n+1} = x_n + n - 1 \)

Also given:
\( x_1 = -1 \)

Let's compute a few initial terms to identify the pattern:

• \( x_2 = x_1 + (1 - 1) = -1 + 0 = -1 \)
• \( x_3 = x_2 + (2 - 1) = -1 + 1 = 0 \)
• \( x_4 = x_3 + (3 - 1) = 0 + 2 = 2 \)
• \( x_5 = x_4 + (4 - 1) = 2 + 3 = 5 \)

Now, observe the pattern more generally:

\[ x_2 = x_1 + (1 - 1) \\ x_3 = x_2 + (2 - 1) = x_1 + (1 - 1) + (2 - 1) \\ x_4 = x_3 + (3 - 1) = x_1 + (1 - 1) + (2 - 1) + (3 - 1) \]

Therefore, in general: \[ x_{n} = x_1 + \sum_{k=1}^{n-1}(k - 1) \]

Simplifying the sum: \[ \sum_{k=1}^{n-1}(k - 1) = \sum_{k=1}^{n-1}k - \sum_{k=1}^{n-1}1 = \frac{(n-1)n}{2} - (n - 1) \] \[ = (n - 1)\left( \frac{n}{2} - 1 \right) = \frac{(n - 1)(n - 2)}{2} \]

So we have: \[ x_n = x_1 + \frac{(n - 1)(n - 2)}{2} \]

Now, substitute \( x_1 = -1 \) and \( n = 100 \): \[ x_{100} = -1 + \frac{(99)(98)}{2} = -1 + 4851 = \mathbf{4850} \]

Final Answer: \( \boxed{x_{100} = 4850} \)