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If $x_{1} = \;– 1$ and $x_{m} = x_{m+1} + (m + 1)$ for every positive integer $m,$ then $x_{100}$ equals

1. $– 5151$
2. $– 5150$
3. $– 5051$
4. $– 5050$

Given that,

• $x_{1} = \;– 1$
• $x_{m} = x_{m+1} + (m+1); \; m>0 \quad \longrightarrow (1)$

We can re-write the equation $(1).$

$x_{m+1} = x_{m} – (m+1)$

Put $m = 1,2,3,4,5, \dots$

Then,

• $x_{2} = x_{1} – 2 \Rightarrow x_{2} = – 1 – 2$
• $x_{3} = x_{2} – 3 \Rightarrow x_{3} = – 1 – 2 – 3$
• $x_{4} = x_{3} – 4 \Rightarrow x_{4} = – 1 – 2 – 3 – 4$
• $x_{5} = x_{4} – 5 \Rightarrow x_{5} = – 1 – 2 – 3 – 4 – 5$
• $\vdots \;\; \vdots \;\; \vdots\;\; \vdots\;\; \vdots$

So, $x_{100} = \;– 1 – 2 – 3 – 4 – 5 – \;\dots\; – 100$

$\Rightarrow x_{100} =\; – ( 1 + 2 + 3 + 4 + 5 + \dots + 100 )$

$\Rightarrow x_{100} = \;– \left[ \frac{100(101)}{2} \right]$

$\Rightarrow x_{100} = \;– 50 \times 101$

$\Rightarrow \boxed{ x_{100} =\; – 5050}$

$\therefore$ The value of $x_{100}$ is $\;– 5050.$

Correct Answer$: \text{D}$

$\textsf{PS:}$ Sum of first $n$ natural numbers $= 1 + 2 + 3 + 4 + 5 + \dots + n = \dfrac{n(n+1)}{2}.$

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