A recursive relation: \[ x_{m+1} = x_m - (m + 1) \] with base value: \[ x_1 = -1 \]
Using the recurrence:
From the pattern, we see: \[ x_n = - (1 + 2 + 3 + \dots + n) = -\frac{n(n+1)}{2} \]
To find: \[ x_{100} = -\frac{100 \times 101}{2} = -5050 \]
\(\boxed{-5050}\)
\[ x_1 = -1, \quad \text{and} \quad x_m = x_{m+1} + (m+1) \] Rearranging the recursive formula: \[ x_{m+1} = x_m - (m+1) \]
Let’s compute the first few values:
Clearly, the pattern becomes: \[ x_n = - (1 + 2 + 3 + \dots + n) \] Which is the negative of the sum of the first \( n \) natural numbers.
\[ x_n = -\frac{n(n+1)}{2} \]
\[ x_{100} = -\frac{100(101)}{2} = -50 \times 101 = \boxed{-5050} \]
The correct option is (D): \[ \boxed{-5050} \]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
When $10^{100}$ is divided by 7, the remainder is ?