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

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4850 — Easy Unroll the recurrence $x_{n+1}=x_n+(n-1)$ by summing the increments from $x_1$ up to $x_{100}$. The increments form $0,1,2,\dots,98$, a simple arithmetic-series sum. 1 List the increments. Going from $x_n$ to $x_{n+1}$ adds $n-1$. $$\begin{aligned} &x_2 = x_1 + 0\ &x_3 = x_2 + 1\ &x_4 = x_3 + 2,\ \dots \end{aligned}$$ 2 Telescope up to $x_{100}$. The increments added are $0+1+2+\cdots+98$. $$\begin{aligned} &x_{100} = x_1 + (0+1+2+\cdots+98)\ &\Rightarrow\ x_{100} = x_1 + \frac{98\cdot 9

CAT 2021 Slot 3 — QA Question 17

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