Value of x100 in recursive sequence

CAT 2020 Slot 3 · QA · Medium · Inequalities and polynomials

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If x1=−1x_1 = -1 and xm=xm+1+(m+1)x_m = x_{m+1} + (m + 1) for every positive integer mm, then x100x_{100} equals

  1. A.

    -5050

  2. B.

    -5051

  3. C.

    -5150

  4. D.

    -5151

Answer

A

Explanation

Given xm−xm+1=m+1x_m - x_{m+1} = m + 1. Summing from m=1m = 1 to 9999: (x1−x2)+(x2−x3)+⋯+(x99−x100)=2+3+⋯+100(x_1 - x_2) + (x_2 - x_3) + \dots + (x_{99} - x_{100}) = 2 + 3 + \dots + 100 x1−x100=100×1012−1=5050−1=5049x_1 - x_{100} = \frac{100 \times 101}{2} - 1 = 5050 - 1 = 5049. Since x1=−1x_1 = -1, we have −1−x100=5049  ⟹  x100=−5050-1 - x_{100} = 5049 \implies x_{100} = -5050.

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