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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 $
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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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