Consider a sequence of real numbers x1, x2, x3, … such that xn+1 = xn + n – 1 for all n ≥ 1. If x1 = -1 then x100
Explanation:
By substituting x = 1 or 2 or 3 and so on, we get
x2 = x1 + 1 – 1 = x1 + 0 x3 = x2 + 2 – 1 = x2 + 1 = x1 + 0 + 1 x4 = x3 + 3 – 1 = x3 + 2 = x1 + 0 + 1 + 2 x5 = x4 + 4 – 1 = x4 + 3 = x1 + 0 + 1 + 2 + 3 … xn = xn-1 + 0 + 1 + 2 + … + (n – 2)
⇒ x100 = x1 + 0 + 1 + 2 + … + 98 = x1 + (98 × 99)/2
⇒ x100 = -1 + 4851 = 4850
Hence, option (a).
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